IN   MEMORIAM 
FLORIAN  CAJORl 


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Digitized  by  the  Internet  Archive 

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http://www.archive.org/details/elementsofanalytOOwoodrich 


THE 


ELEMENTS 

or 

ANALYTICAL    MECHANICS 

SOLIDS  AND   FLUIDS. 


DeVOLSON   wood,  a.m.,  C.E., 

WRonsson  of  kathekatics  and  mkcSanics  in  stkvkns   institute  of  tkchnologtj 

AUTHOR    OF     "BE8I8TANCE     OF     MATERIALS;"     "ROOFS     AND     BRIDGES;"      EEVISKD 
XOmOK  OF  "  MAHAjr's  CITLL  XNOUrXXKUfO,"  AUD  "  KI.SKXNTASY  lUCHASICS." 


FIFTH  EDITION. 


BKVISKD,  CORBKCTBO,  AMD    ENLAROBD. 


NEW  YORK : 

JOHN  WILEY  &  SONS, 

15  AsTOR  Place. 

1887. 


OorTBIOHTED,    1876,    »T 

devolson  wood. 


PREFACE, 


The  plan  of  this  edition  is  the  same  as  the  former  one.  It 
is  designed  especially  for  students  who  are  beginning  the  study 
of  Analytical  Mechanics,  and  is  preparatory  to  the  higher  works 
upon  the  same  subject,  and  to  Analytical  Physics  and  Astro- 
nomy. The  Calculus  is  freely  used.  I  have  sought  to  present 
the  subject  in  such  a  manner  as  to  familiarize  the  student  with 
analytical  processes.  For  this  reason  the  solutions  of  problems 
have  been  treated  as  applications  of  general  formulas.  The 
solution  by  this  method  is  often  more  lengthy  than  by  special 
methods;  still,  it  has  advantages  over  the  latter,  because  it 
establishes  a  uniformity  in  the  process. 

My  experience  has  shown  the  importance  of  applying  the 
fundamental  equations  to  a  great  variety  of  problems.  I  have, 
therefore,  in  Article  24,  and  Chapters  lY.  and  X.,  given  a  large 
number  and  a  considerable  variety  of  problems  to  be  solved  by 
the  general  equations  under  which  they  respectively  fall. 

In  the  revision  I  have  been  aided  not  only  by  my  own  expe- 
rience with  the  use  of  the  former  edition  in  the  class-room,  but 
also  by  the  friendly  advice  and  criticism  of  several  professors 
of  colleges  who  have  used  the  work.  The  result  has  been  that 
several  pages  have  been  rewritten,  some  definitions  changed, 
and  the  typographical  errors  corrected.  Several  new  pages  in 
the  latter  part  of  the  work  have  been  added.  I  am  especially 
indebted  to  Professor  E.  T.  Quimby,  of  Dartmouth  College, 
Hanover,  N.  H.,  for  his  valuable  suggestions  and  for  assistance 
in  reading  the  final  proofs. 

The  nature  of  force  remains  as  much  a  mystery  as  it  waa 


rr:r»rvit  CDCzyr^ 


IV  PilEPACB. 

when  its  principles  were  first  i-ecognized.  Of  its  essential 
nature  we  shall  probably  remain  forever  in  ignorance.  We  can 
only  deal  with  the  Xaws  of  its  action.  These  laws  are  deter- 
mined by  observing  the  effects  produced  by  a  force.  Force  is 
the  cause  of  an  action  in  the  physical  world.  The  results  of 
the  action  may  be  numerous  and  varied.  Thus,  force  may  pro- 
duce pressure,  tension,  cohesion,  adhesion,  motion,  affinity,  po- 
larity, .electricity,  etc.  Or,  to  speak  more  properly,  since  force 
may  be  transmuted  from  one  state  to  another,  we  would  say 
that  the  above  terms  are  names  for  the  different  manifestations 
of  force. 

The  question  "  what  is  the  correct  measure  of  force"  has  taken 
different  phases  at  different  times.  During  the  last  century  it 
was  contended  by  some  that  momentum  {Mv)  was  tlie  correct 
measure,  while  others  contended  that  it  should  be  the  work  which 
it  can  do  in  a  unit  of  time  {iMir^).  But  as  one  has  happily 
expressed  it,  "  theirs  was  only  a  war  of  words ; "  for  the  real 
measure  of  force  enters  only  as  a  factor  in  the  expressions. 
Thus,  if  /^  be  a  constant  force,  the  value  of  the  momentum  is 
J^t,  see  page  51,  and  of  the  work  7^6%  see  page  45.  At  the  pre- 
sent day  some  contend  that  the  only  measure  of  force  is  the 
motion  which  it  produces,  or  would  produce,  in  a  unit  of  time. 
This  is  called  the  absolute  measure,  and  the  absolute  unit  op 
FOKCE  is  the  velocity  which  the  force  produces,  or  would  jyro- 
dttce,  in  a  unit  of  mass  in  a  unit  of  time  if  it  acted  during 
the  unit  with  the  intensity  which  it  had  at  the  instant  con- 
sidered. If  the  intensity  of  the  force  were  constant,  the  velo- 
city which  it  produced  at  the  end  of  the  unit  of  time  would 
be  the  required  velocity.  Hence,  the  absolute  measure  of  any 
force  acting  on  any  mass  is  the  product  of  the  mass  into  the 
acceleration;  and  is  the  second  member  of  equation  (21). 
This  is  a  correct  measure,  and  is  accepted  as  such  by  all  writers 
on  mechanics. 

But  those  who  contend  that  this  is  the  only  measure,  neces- 
sarily deny  that  weight,  or  more  generally  pounds,  is  a  mea- 
sure.    1  contend  'dL\2^\, pounds  is  a  measure  of  the  intensity  of  8 


PREFACE.  ▼ 

force  both  statically  and  dynamically.  Many  authoi-s  maintain 
the  same  position.  Indeed,  it  is  probable  that  the  position 
which  I  have  taken  can  be  deduced  from  an}^  standard  work  on 
mechanics ;  but  in  some  it  is  left  to  inference.  Thus,  in 
Smith's  Mechanics^  P^^^  Ij  ^ve  find  this  terse  and  correct  defi- 
nition, "  The  intensity  of  a  force  may  be  measured,  statically, 
by  the  pressure  it  will  produce ;  dynamically,  by  the  quantity 
of  motion  it  will  produce."  I  say  this  is  correct,  but  I  will  add 
that  the  intensity  of  a  force  which  produces  a  given  motion  is 
also  measured  by  a  pressure,  or  by  something  equivalent  to  a 
pressure,  or  to  a  pull.  To  those  who  will  look  at  it  analyti- 
cally, it  is  only  necessary  to  say  that  the  first  member  of  equa- 
tion (21)  is  measured  in  pounds.  If  we  know  the  absolute 
measure,  we  may  easily  find  its  value  in  pounda, 

The  pound  here  i-ef  erred  to  is  the  result  of  the  action  of  gra- 
vity upon  a  certain  quantity  of  matter.  The  amount  of  matter 
having  been  fixed,  either  by  a  legal  enactment  or  by  common 
consent,  and  declared  to  be  one  pound  at  a  certain  place,  its 
weight,  as  determined  by  a  standai-d  spring-balance  at  any  other 
place,  becomes  a  measure  of  the  foi-ce  of  gravity  as  compared 
with  the  fixed  place.  This  standard  spring-balance  may  meas- 
ure the  intensity  in  pounds  of  any  other  force,  whether  the 
body  upon  which  the  force  acts  be  at  rest  or  in  motion.  If  a 
perfectly  free  body  were  placed  in  a  hollow  space  at  the  centre 
of  the  earth,  at  which  place  it  would  be  devoid  of  weight,  and 
pulled  or  pushed  by  a  constant  force,  whose  intensity,  measured 
by  a  standard  spring-balance,  equaled  the  weight  of  an  equal 
body  on  the  surface  of  the  earth,  then  would  its  motion  be  the 
same  as  that  of  a  falling  body.  See  page  24,  Problem  7.  In 
the  forces  of  nature  producing  motion,  there  being  no  visible 
connection  between  the  point  of  action  of  the  force  and  the 
body  upon  which  it  acts,  we  are  unable  to  wti^h  their  intensity 
except  by  calculation.  If  the  absolute  measure  is  known,  the 
jpounds  of  intensity  may  be  computed.  The  absolute  measure 
of  the  force  of  gravity  on  a  mass  m  is  m^,  and  the  weight  of 
the  body  being  TF,  we  have  W—  rng.  The  sun  acts  upon  the 
earth  with  a  force  which  may  be  expressed  by  the  absolute 


VI  PREFACE. 

measure,  and  also  by  a  certain  nnmber  of  pounds  of  force. 
More  than  half  of  the  examples  in  Article  24  involve  an 
equality  between  pouTids  of  intensity  and  the  absolute  measure 
of  the  force.  The  fact  is,  that,  in  case  of  motion,  these  quanti- 
ties are  co- relative.  Since,  then,  it  is  correct  to  use  the  term 
pound  as  the  measure  of  the  intensity  of  a  force  whether  the 
body  be  at  rest  or  in  motion,  and  since  it  is  in  common  use,  and 
the  student  is  familiar  with  it,  I  prefer  to  consider  a  force  as 
measured  by  a  certain  number  of  pounds.  See  Article  9.  It 
is  more  simple,  containing  as  it  does  only  one  element,  than 
the  absolute  measure,  which  contains  three  elements — ^mass, 
velocity,  and  time. 

There  is  another  advantage  in  thus  measuring  force.  Stu- 
dents frequently,  and  in  some  cases  writers,  use  the  expressions, 
"  quantity  of  force,"  "  amount  of  force,"  "  force  of  a  blow,"  etc., 
when  they  mean  (or  should  mean)  momentum,  or  work,  or  via 
viva.  In  such  cases  an  attempt  to  answer  the  question  "how 
many  pounds  of  force  "  would  show  at  once  that  tiie  quantity  re- 
ferred to  was  T\otfo7'ce. 

So  much  ambiguity,  or  at  least  indeiiniteness,  has  arisen  in 
regard  to  the  term  force,  that  I  have  rejected  the  terms  "  Im- 
pulsive Force  "  and  "  Instantaneous  Force,"  and  used  the  term 
"  Impulse  "  instead  of  them.  "We  know  nothing  of  an  instan- 
taneous force,  that  is,  one  which  requires  no  time  for  its  action. 
I  also  reject  the  expression /c>r(?6  of  inertia,  I  do  not  believe 
that  inertia  is  ^  force.  To  the  question  "  The  inertia  of  a  body 
is  how  many  pounds  of  force  "  there  is  no  answer. 

The  term  moinent  of  inertia  has  no  physical  representation. 
The  nearest  approach  to  it  is  in  the  expression  for  the  vis  viva 
of  a  rotating  body.  In  such  problems  the  moment  of  inertia 
forms  an  important  factor.  The  energy  of  a  rotating  body  hav- 
ing a  constant  angular  velocity  is  directly  proportional  tp  its 
moment  of  inertia  in  reference  to  its  axis  of  rotation.  See  page 
202.  But  motion  is  not  necessary  for  its  existence.  See  page 
165.     The  expression  appears  in  the  discussion  of  numerouB 


PREFACE.  vn 

statical  problems,  such  as  the  flexure  of  a  beam,  the  centre  of 
pressure  of  a  fluid,  the  centre  of  gravity  of  certain  solids,  etc. 
It  is  not  the  moment  of  a  moment,  although  it  may  be  so  con- 
strued as  to  appear  to  be  of  thsit  form.  Some  other  term  might 
be  more  appropriate.  Even  the  expression  moment  of  the  mass 
would  be  less  objectionable. 

The  subjects  of  Centrifugal  Force  and  Unbalanced  Force 
have  been  discussed  of  late  in  Engineering.  Some  assert  that 
there  is  no  such  thing  as  a  centrifugal  force.  Much  unprofita- 
ble discussion  may  be  avoided  by  strictly  defining  the  terms 
used.  If  it  is  defined  to  be  a  force  equal  and  opposite  to  the 
deflecting  force,  it  will,  at  least,  have  an  ideal  existence,  just  as 
the  resultant  in  statical  problems  has  an  ideal  existence.  But 
the  vital  question  is,  is  the  centrifugal  force  active  when  the 
deflecting  force  acts  \  Or,  in  other  words,  do  both  act  upon 
the  body  at  the  same  time  ?  It  seems,  however,  quite  evident 
that  if  both  acted  upon  the  body  at  the  same  time  they  would 
neutralize  each  other,  and  the  body  would  move  in  a  straight 
line.  Hence,  in  the  movement  of  the  planets,  or  of  any  free 
rotating  body,  there  is  no  centrifugal  force.  But  in  the  case  of 
a  locomotive  running  around  a  curve  there  may  be  both  cen- 
tripetal and  centrifugal  forces ;  the  former  acting  against  the 
locomotive  to  force  it  away  from  a  tangent  to  the  track ;  the 
latter,  against  the  track,  tending  to  force  it  outward.  Wher- 
ever the  force  is  conceived  to  act,  whether  just  between  the 
rail  and  wheel  or  at  some  other  point,  it  is  evident  that  both  do 
not  act  upon  the  same  body. 

Similarly  in  regard  to  the  unbalanced  force.  It  is  a  conveni 
ent  term  to  use,  but,  in  a  strict  sense,  an  unbalanced  force  doeb 
not  exist ;  for  action  and  reaction  are  equal  and  opposite.  But 
in  reference  to  a  particular  body,  all  other  conditions  being 
ignored,  the  force  may  be  unbalanced.  Thus,  when  a  ball  ia 
fired  from  a  cannon,  the  force  of  the  powder,  considered  in  the 
direction  of  the  motion  of  the  ball  only,  is  unbalanced ;  but  the 
powder  exerts  an  equal  force  in  the  opposite  direction,  and  in 
that  sense  also  is  unbalanced.     But  when  the  entire  effect  of  the 


Vlll  PREFACE. 

force  in  all  directions  is  considered,  the  algebraic  resultant  is 
zero.  In  other  words,  the  centre  of  gravity  of  the  system,  for 
forces  acting  between  its  integrant  parts,  remains  constant. 

These  are  some  of  the  fundamental  questions  which  will  arise 
in  the  mind  of  the  student  as  he  studies  the  subject.  Fortu- 
nately, it  is  not  necessary  for  him  to  settle  them  beyond  the 
question  of  a  doubt  before  he  proceeds  with  the  subject.  On 
some  of  these  points  scholars,  who  have  made  the  subject  a 
specialty,  differ ;  and  it  is  only  after  a  careful  consideration  of 
the  points  involved  that  one  can  take  an  intelligent  position  in 
regard  to  them. 

DeYolson  Wood. 

HOBOKEN,  August,  18T7. 


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GREEK  ALPHABET. 


Letters. 

Names. 

Letters. 

Namoik               i 

Nu        W 

A  a 

Alpha  (> 

JV  I. 

B  13 

Beta    J^ 

s  f 

Xi         X 

ry 

Gamma  ^ 
Delta  0^1 

0  0 

Omicron  6 

A  8 

n  TT 

Pi 

E    € 

Epsilon  Ji. 

^P 

Kho     r 

Zf 

Zeta     5^ 
Eta     ^ 

2*  <r  9 

Sigma       vS 

Hv 

Tt 

Tail             :_ 

e  ^(9 

Theta    -^ 

r  V 

Upsilon       Y 

I  I 

I5ta       ^ 

<^g) 

Phi             ^ 
Chi     QJt^^ 

K   K 

Kappa     k 
Tiambda   <.^ 

Xx 

A  \ 

F./r 

Psi      bs 

M   fJL 

Mu            1  v^ 

/2  a> 

Omega       O" 

r 


TABLE  OF  CO:S TENTS. 


CHAPTER  I. 


DEFINITIONS  AND  THB  LAWS  OP  FORCES  WniCH  ACT   ALONG    A 
STRAIGHT  LINE. 

ABTICLBS  tAOB 

1-14 — Definitions. 1 

15_oo_  Velocity  ;  Gravity;  Mass 6 

21-23— Force;  Mass;  Density 16 

24 — Elxamples  of  Accelerated  Motion 21 

25-27 — Work  ;  Energy  ;  Momentum 44 

28-33— Impact 53 

34-36  -  Statics  ;   Power;  Inertia .^ 58 

87-38 — Newton's  Laws  of  Motion.    Eccentric  Impact 62 


CHAPTER  II. 

COMPOSITION  AND  RESOLUTION  OP  FORCES. 

39-42 — Conspiring  Forces 65 

43-46 — Composition  of  Forces 65 

47 — Resolution  on  Three  Axes 70 

48-49 — Constrained  Equilibrium  of  a  Particle 71 

50-60— Statical  Moments 79 

Examples 84 


CHAPTER  III. 

PARALLEL  FORCES. 

61-63— Resultant 86 

64— Moments  of  Parallel  Forces 87 

65-68— Statical  Couples 88 

69-  74— Centre  of  Gravity  of  Bodies 92 

75 — Centrobaric  Method 103 

77-81— General  Properties  of  the  Centre  of  Gravity 108 

82— Centre  of  Mass 110 


XU  TABLE  OF  CONTENTS. 


CHAPTER  IV. 

NONCONCURRENT  FORCES. 
ABTicLKS  piaa 

83-  85— General  Equations. Ill 

86— Problems 115 

CHAPTER  V. 

STRESSES. 

'a7_  90— Stresses  Resolved 142 

91-  93 — Shearing  Stresses — Notation 144 

93_  94— Resultant  Stress 140 

95-  96 — Discussion 158 

97-  98 — Conjugate  Stresses.     General  Problem 156 


CHAPTER  VL 

VIRTUAL  VELOCITIES. 

100 — Concurring  Forces 159 

101 — Nonco»curring  Forces 160 

Examples 161 


CHAPTER  VII. 

MOMENT  OF  INERTIA. 

102-104— Examples.     Formula  of  Reduction 165 

105 — Relation  between  Rectangular  Moments : 169 

106— Moments  of  Inertia  of  Solids 172 

107— Radius  of  Gyration 173 


CHAPTER  VIII. 

MOTION  OF  A  FREE  PARTICLE. 

108 — General  Equations 175 

109_Velocity  and  Living  Force 176 

110-117— Central  Forces 180 


CHAPTER  IX. 

CONSTRAINED  MOTION  OF  A  PARTICLE. 

118-121— General  Equations 191 

122-124— Centrifugal  Force  on  the  Earth 193 


TABLE  OF  CONTENTS.  XUl 

CHAPTER  X. 

BOTATION  OF  A  BODY  WHEN  THB  FORCES  ABE  IN  A  PLANE. 

ABTICLB8  PAOK 

125-131— General  Equations 200 

132— Reduced  Mass 204 

133-135— Spontaneous  Aids ;  Centre  of  Percussion  ;  Aids  of  Instantaneous 

Rotation 209 

Examples 210 

136_1 38— Compound  Pendulum.    Captain  Eater's  Experiments 218 

139-142^— Eliipticity  of  the  Earth.     Torsion  Pendulum 221 

143— Density  of  the  Earth 224 

144_146_Problems 227 

CHAPTER  XI. 

GENERAL  EQUATIONS  OF  MOTION. 

147— D'Alembert's  Principle 230 

148 — General  Equations 231 

149— Conservation  of  the  Centre  of  Gravity 235 

150 — Conservation  of  Areas 235 

151 — Conservation  of  Energy 236 

152— Composition  of  Angular  Velocities 240 

153— Moments  of  Rotation  of  the  Centre 242 

154— Motion  of  the  Centre  of  a  Body 242 

155 — Motion  of  Rotation  of  the  Centre 243 

156— Euler's  Equations 243 

157 — Principal  Axes    244 

158 — No  Strain  on  Principal  Axes 246 

159 — Relation  betvsreen  the  Fixed  and  Principal  Axes 247 

160 — Relations  between  Angular  Velocities 247 

161 — Axis  of  Spontaneous  Rotation 252 

162— Relation  between  Spontaneous  and  Central  Axes 255 

164 — Centre  of  Percussion 263 

165 — Conservation  of  the  Centre  of  the  Mass 264 

166— Conservation  of  the  Moment  of  the  Momentum 264 

167— The  Invariable  Plane 266 

168— Mutual  Action  between  Particles 267 

169 — Attraction  of  a  Homogeneous  Sphere 268 

169a-Mass  and  Stress 271 

1696-Repulsive  Forces 273 

ITO— Principle  of  Least  Action 273 

170a-Gauss'  Theorem  of  Least  Constraint 274 


Xiv  TABLE. OF  CONTENTS. 


CHAPTER  XII. 

MECHANICS  OP  FLUIDS. 
ARTICLES  PAGE 

171— Introductory 275 

173— Definitions 276 

173— Elastic  and  Non-Elastic  Fluids 273 

174_Equality  of  Pressures '. 276 

175— Perfect  Fluid  at  Rest 277 

176— The  Law  of  Equal  Transmission 277 

177 — Pressure  on  Base  of  Prismatic  Vessel 278 

^  178— Pressure  on  Base  of  any  Vessel 278 

_  179— Static  Head 279 

180— Free  Surface v 279 

181 — Pressure  on  Submerged  Surface 280 

182— Resolved  Pressures 281 

183— Resultant  Pressure 281 

^  184— Centre  of  Pressure 282 

185— Condition  of  Flotation 285 

186— Depth  of  Flotation 286 

187— Specific  Gravity 288 

188— Examples 289 

189— Fluid  Motion.     Definitions 291 

190— Bernouilli's  Theorem 292 

191— Discussion  of 294 

192 — Graphic  Representation 295 

193 — Flowing  through  Varying  Sections.      Examples 296 

194— Velocity  of  Discharge  from  Small  Orifice 300 

195 — Discharge  from  Large  Orifice 301 

196— Vertical  Orifices.     Examples 302 

197— Conical  Adjutages 307 

198— Reaction  of  Fluids 310 

199— Centrifugal  Force 310 

200— Resolved  Centrifugal  Force 311 

202— Pressure  Due  to  Discharge  from  the  Side  of  a  Vessel 314 

203— Pressure  Due  to  a  Discharge  Vertically  Upward 315 

205 — Pressure  of  a  Stream  Impinging  Normally  Against  a  Plane  Sur- 
face   316 

206— Case  of  Cup  Vane 316 

207— Case  of  Bent  Tube 317 

208— Case  of  Inclined  Surface '.319 

209— Remark.     Examples •. 319 

210— Case  of  Impinged  Surface  in  Motion 322 

211— Work  Done  on  Vane 323 

212— Energy  Imparted  to  a  Vane 324 


TABLE   OF  CONTENTS.  XV 

ABTIOLES  PAGE 

213— Expression  for  EflBciency 325 

214— Resultant  Pressure  on  a  Vane 325 

215— General  Case 326 

216— Discussion 327 

217— Hydraulic  Motors 832 

218— Case  of  a  Single  Vane 332 

220— The  Undershot  Wheel 383 

222— The  Poncelet  Wheel 336 

223— The  Breast  Wheel 337 

224^The  Overshot  Wheel.    Examples 337 

225— Effective  Head 342 

226— The  Jet  Propeller 843 

227— Barker's  Mill 343 

228 — Pressure  Due  to  Centrifugal  Force.      Examples 348 

229— Definitions  of  Turbines v 350 

230— Analysis  of  Turbines 351 

231— Special  Cases 355 

232— General  Case.     Examples 359 

;  233 — Resistance  Due  to  Sudden  Enlargements 862 

(  234 — Resistance  in  Long  Pipes 363 

235 — General  Case  of  Flow  in  Long  Pipes.     Examples 864 

236— Gases 369 

237— Boyle's  Law 370 

238— Tension  of  Gas  in  a  Vertical  Prism 371 

289— Numerical  Values  in  Regard  to  Air 372 

240— Tension  of  the  Atmosphere  Gravity  Variable 372 

241— Heights  by  Barometer 374 

242— Gay  Lussac's  Law 375 

243— Absolute  Zero 376 

244— Heat  a  Form  of  Energy 377 

245— Thermal  Unit 878 

246— Specific  Heat 878 

247— Dynamic  Specific  Heat 379 

248— The  Adiabatic  Curve.     Examples 379 

249— Velocity  of  a  Wave  in  an  Elastic  Medium 882 

250— Value  of  ;^ 887 

251— Remarks.     Examples 387 

252— Velocity  of  Discharge  of  Gases 388 

254— Weight  of  Gas  Discharged 390 


APPENDIX  I. 
Solutions  of  Problems 891-463 

APPENDIX  II. 
The  Potential 467-470 


ANALYTICAL   MECHANICS. 


CHAPTER  I. 

DEFINITIONS,  AND   PRINCIPLES  OF  ACTION   OF  A   SINGLE   FORCE,  AND 
OF   FORCES    ACTING   ALONG    THE    SAME   LINE. 

1.  Mechanics  treats  of  the  laws  of  forces,  and  the  equi- 
librium aud  motion  of  bodies  under  the  action  of  forces.  It 
has  two  grand  divisions,  Dynamics  and  Statics. 

2.  Dynamics  treats  of  the  motion  of  material  bodies,  and 
the  laws  of  the  forces  which  govern  their  motion. 

3.  Statics  treats  of  the  conditions  of  the  equilibrium  of 
bodies  under  the  action  of  forces. 

There  are  many  subdivisions  of  the  subject,  such  as  Hydrodynamics,  Hy- 
drostatics, Pneumatics,  Thermodynamics,  Molecular  Mechanics,  etc.  That 
part  of  mechanics  which  treats  of  the  relative  motion  of  bodies  which  are  so 
connected  that  one  drives  the  other,  such  as  wheels,  pulleys,  links,  etc.,  in 
machinery,  is  called  Cinematics.  The  motion  in  this  case  is  independent  of 
the  intensity  of  the  force  which  produces  the  motion. 

Theoretic  Mechanics  treats  of  the  effect  of  forces  applied  to  material  points 
or  particles  regarded  as  without  weight  or  magnitude.  Somatology  is  the 
application  of  theoretic  mechanics  to  bodies  of  definite  form  and  magnitude. 

4.  Matter  is  that  which  receives  and  transmits  force.  In 
a  physical  sense  it  possesses  extension,  divisibility,  and  impene- 
trability. 

Matter  is  not  confined  to  the  gross  materials  which  we  see  and  handle,  but 
includes  those  substances  by  which  sound,  heat,  light,  and  electricity  are 
transmitted. 

It  is  unnecessary  in  this  connection  to  consider  those  refined  speculations  by 
which  it  is  sought  to  determine  the  essential  nature  of  matter.  According  to 
some  of  these  speculations,  matter  does  not  exist,  but  is  only  a  conception. 


2  DEFINITIONS.  [6-6.] 

According  to  this  view,  bodies  are  forces,  within  the  limit  of  which  the  attrac- 
tive exceed  the  repulsive  ones,  and  at  the  limits  of  which  they  are  equal  to 
each  other. 

But  observation,  long  continued,  teaches  practicaDy  that  matter  is  inert, 
that  it  has  no  power  within  itself  to  change  its  condition  in  regard  to  rest  or 
motion ;  that  when  in  motion  it  cannot  change  its  rate  of  motion,  nor  be 
brought  to  rest  without  an  external  cause,  and  this  cause  we  call  FOilCE. 
One  also  learns  from  observation  that  matter  will  transmit  a  force,  as  for 
instance  a  pxill  applied  at  one  end  of  a  bar  or  rope  is  transmitted  to  the  other 
end ;  also  a  moving  body  carries  the  effect  of  a  force  from  one  place  to  another. 

5.  A  Body  is  a  definite  portion  of  matter.  A  particle  is 
an  infinitesimal  portion  of  a  body,  and  is  treated  geometrically 
as  a  point.  A  molecule  is  composed  of  several  particles.  An 
atom  is  an  indivisible  particle. 

6.  Force  is  that  which  tends  to  change  the  state  of  a  body 
in  regard  to  rest  or  motion.  It  moves  or  tends  to  move  a  body, 
or  change  its  rate  of  motion. 

We  know  nothing  of  the  essential  nature  of  force.  "We  deal 
only  with  the  laws  of  its  action.  These  laws  are  deduced  by 
observations  upon  the  effects  of  forces,  and  on  the  hypothesis 
that  action  and  reaction  are  equal  and  ojpposite ;  or,  in  other 
words,  that  the  effect  equals  the  cause.  In  this  way  we  find 
that  forces  have  different  intensities,  and  that  a  relation  may 
be  established  between  them.  It  is  necessary,  therefore,  to 
establish  a  unit.  This  may  be  assumed  as  the  effect  of  any 
known  force,  or  a  multiple  part  thereof.  The  effect  of  all 
known  forces  is  to  produce  a  pull,  or  push,  or  their  equivalents, 
and  may  be  measured  by  pounds,  or  by  something  equivalent. 
The  force  of  gravity  causes  the  weight  of  bodies,  and  this  is 
measured  by  pounds.  We  therefore  assume  that  a  standard 
POUND  is  the  UNIT  of  force. 

The  standard  pound  is  established  by  a  legal  enactment,  and 
has  been  so  fixed  that  a  cubic  foot  of  distilled  w^ater  at  the 
level  of  the  sea,  at  latitude  45  degrees,  at  a  temperature  of  62 
degrees  Fahrenheit,  with  the  barometer  at  30  inches,  will  weigli 
about  62.4  pounds  avoirdupois. 

The  English  standard  pound  was  originally  5,760  troy  grains.  The  grain 
was  the  weight  of  a  certain  piece  of  brass  which  was  deposited  with  the  clerk 
of  the  House  of  Commons.  This  was  destroyed  at  tho  time  of  the  burning  of 
the  House  of  Commons  in  1884,  after  which  it  was  decided  that  the  legal 


[7.]  STANDARD  UNITS.  O 

pound  should  be  the  weight  of  a  certain  piece  of  platinum,  weighing  7,001' 
grains.  This  is  known  as  the  avoirdupois  pound,  and  the  troy  pound  ceased 
to  be  the  legal  standard,  although  both  have  remained  in  common  use. 

The  legal  standard  pound  in  the  United  States  is  a  copy  of  the  English  troy 
pound,  and  was  deposited  in  the  United  States  Mint  in  Philadelphia,  in  1827, 
where  it  has  remained.  The  avoirdupois  pound,  or  7,000  grains,  is  used  in 
nearly  all  commercial  transactions.  The  troy  pound  is  a  standard  at  62  de- 
grees Fahrenheit  and  30  inches  of  the  barometer. 

The  weight  of  a  cubic  inch  of  water  at  its  maximum  density,  as  accepted 
by  the  Bureau  of  Weights  and  Measures  of  the  United  States,  is  stated  by  Mr. 
Hasler,  in  a  report  to  the  Secretary  of  the  Treasury,  1842,  to  be  252.7453 
grains.  Mr.  Hasler  determined  the  temperature  at  which  water  has  a  maxi- 
mum density,  at  39.83  degrees  Fahrenheit,  but  Playfair  and  Joule  determined 
it  to  be  39.101°  F. 

The  exact  determination  of  the  equivalent  values  of  the  units  is  very  diffi- 
cult, and  has  been  the  subject  of  much  scientific  investigation. — (See  The  Me- 
tric System,  by  F.  A.  P.  Barnard,  LL.D.,  New  York,  1872.) 

When  a  quantity  can  be  measured  directly,  the  unit  is  generally  of  the  same 
quality  as  the  thing  to  be  measured  :  thus,  the  unit  of  time  is  time,  as  a  day 
or  second  ;  the  unit  of  length  is  length,  as  one  inch,  foot,  yard,  or  metre  ;  the 
unit  of  volume  is  volume,  as  one  cubic  foot ;  the  unit  of  money  is  money  ;  of 
weight  is  weight ;  of  momentum  is  momentum ;  of  work  is  work,  etc. 

When  dissimilar  quantities  are  used  to  measure  each  other  a  proportion 
must  be  established  between  them.  It  is  commonly  said  that  ' '  the  arc  mea- 
sures the  angle  at  the  centre,"  but  it  does  not  do  it  directly,  since  there  is  no 
ratio  between  them.  The  arc  is  a  linear  quantity,  as  feet  or  yards,  or  a  num- 
ber of  times  the  radius,  while  the  angle  is  the  divergence  of  two  lines,  and  is 
usually  expressed  in  degrees.  But  angles  are  proportional  to  their  subtended 
arcs ;  hence  we  have  an  equality  of  ratios,  or 

angle  mhtended  arc 


unit  angle      arc  which  subtends  the  unit  angle 

and  since  a  semi-circumference,  or  ir,  subtends  an  angle  of  180°,  it  is  easy  from 
the  above  equality  of  ratios  to  determine  any  angle  when  the  arc  is  known,  or 
mce  versd. 

Similarly,  the  intensity  of  heat  is  not  measured  directly,  but  by  its  effect  in 
expanding  liquids  or  metals. 

The  magnetic  force  is  measured  by  its  effect  upon  a  magnetic  needle. 

The  intensities  of  lights  by  the  relative  shadows  produced  by  them. 

Similarly  with  forces,  we  measure  them  by  their  effects. 

Dissimilar  quantities,  between  which  no  proportion  exists,  do  not  measure 
each  other.     Thus  feet  do  not  measure  time,  nor  money  weight. 

Pounds  for  commercial  purposes  represents  quantities  of  matter ;  but  when 
applied  to  forces  it  represents  their  intensities.  In  a  strict  sense,  pounds  does 
not  measure  directly  the  quantity  of  matter,  but  is  always  a  measure  of  a  force. 

7.  The  line  of  action  of  a  force  is  the  line  along  whioh 
tlie  force  moves  or  tends  to  move  a  particle.     If  the  particle  is 


4  FOECE,  SPACE,  TIME.  [^^^  1 

acted  tipoii  by  a  single  force,  the  line  of  action  is  straight. 
This  is  also  called  the  actioTv-line  of  the  force. 

8,  The  point  of  application  of  a  force  is  the  point  at 
which  it  acts.  This  may  be  considered  as  at  any  point  of  its 
action-line.  Thns,  if  a  pull  be  applied  at  one  end  of  a  cord, 
the  effect  at  the  other  end  is  the  same  as  if  applied  at  any 
intermediate  point. 

9,  A  FORCE  is  said  to  be  given  when  the  following  elements 
are  known : — 

1st.  Its  magnitude  {pounds) ; 

2d.  The  dii-ection  of  the  line  along  which  it  acts  {action-line) ; 

3d.  The  direction  along  the  action-line  ( +  or  — ) ;  and, 

4th.  Its  point  of  application. 

A  force  may  be  definitely  represented  by  a  straight  line  ; 
thus,  its  magnitude  may  be  represented  by 

the  length  A  B^  Fig.  1 ;  its  position  by  the    A ^  h 

position  of  the  line  A  B\  its  dii-ection  along  ^^^  j 

the  line  by  the  arrow-head  at  B^  which  indi- 
cates that  the  force  acts  from  A  towards  B ;   and  its  point  of 
application  by  the  end  A, 

10,  Space  is  indefinite  extension,  finite  portions  of  w^iich 
may  be  measured. 

11,  Time  is  duration,  and  may  be  measured. 

Probably  no  definition  will  give  a  better  idea  of  the  abstract  quantities  of 
lime  and  %fpace  than  that  which  is  formed  from  experience. 

12,  A  BODY  is  in  motion  w^hen  it  occupies  successive  portions 
of  space  in  successive  instants  of  time.  In  all  other  cases  it 
is  at  absolute  rest.  Motion  in  reference  to  another  moving 
body  is  relative. 

But  a  body  may  be  at  rest  in  reference  to  surrounding  ob- 
jects and  yet  be  in  motion.  Thus,  many  objects  on  the  surface 
of  the  earth,  such  as  rocks,  trees,  etc.,  may  be  at  rest  in  refer- 
euce  to  objects  around  them,  while  they  move  with  the  earth 
through  space.  Observation  teaches  that  there  is  probably  no 
body  at  absolute  rest  in  the  universe. 

13,  Motion  is  uniform  when  the  body  passes  over  equal 
portions  of  space  in  equal  successive  portions  of  time. 

14,  Yaeiable  motion  is  that  in  which  the  body  passes  o^'ei 
unequal  portions  of  space  in  equal  times. 


[15.]  VELOCITY.  o 

15,  Velocity  is  the  rate  of  motion.  When  the  motion  is 
uniform  it  is  measured  by  the  linear  distance  over  which  a 
body  would  pass  in  a  unit  of  time  ;  and  when  it  is  variable  it 
is  the  distance  over  which  it  would  pass  if  it  moved  with  tlie 
rate  which  it  had  at  the  instant  considered.  The  path  of  n 
moving  particle  is  the  line  which  it  generates. 

For  uniforn^  velocity,  we  have 

in  which 

s  =  the  space  passed  over ; 

t  =  the  time  occupied  in  moving  over  the  space  s ;  and 

V  =  the  velocity. 

JFbr  variable  velocity,  we  have 

EXAMPI.ES.  • 

1.  If  a  particle  moves  uniformly  thirty  feet  in  three  seconds, 
what  is  its  velocity  ? 

2.  If  5  =  at,  what  is  the  velocity  ? 

3.  If  5  =  «^  -f  ht,  what  is  the  velocity  at  the  time  t,  or  at 
the  end  of  the  space  s  ? 

Here 

which  is  the  answer  to  the  first  part.  Find  t  from  the  given 
equation,  and  substitute  in  the  expression  for  v,  and  it  gives 
the  answer  to  the  second  part ;  or 

^=  VW~+Tas. 

4.  If  5  =  4^,  required  the  velocity  at  the  end  of  five  seconds. 

5.  If  Zf?  =  5^,  required  the  velocity  at  the  end  of  ten  sec- 
onds. 

6.  If  «  =  igf,  wliat  is  the  velocity  in  terms  of  the  time  and 
space  ? 

7.  If  at  =  e^'—\,  required  the  velocity  in  terms  of  the  timt 
and  space. 


6 


ANGULAR  VELOCITY. 


[16,17.i 


^^ 


Fig.  2. 


16.  Angular  velocity  is  the  rate  of  angular  movement,  li 
a  particle  moves  around  a  point  having  either  a  constant  or  a 
variable  velocity  along  its  path, 
the  angular  velocity  is  meas- 
ured hy  the  arc  at  a  unites  dis- 
tance which  subtends  the  angle 
swept  over  in  a  unit  of  time  hy 
that  radius  vector  which  passes 
through  the  particle. 

If  5  =  AB  =  the  length  of  the  path  described ; 
V  =  the  velocity  along  the  path  AB  / 
t  =  the  time  of  the  movement ; 
r  =  CB  =  the  radius  vector ; 
6  =  the  circular  arc  at  a  unit's  distance  which  subtends 

the  angle  A  CB  swept  over  by  the  radius  vector  in 

the  time  t ;    and 
to  =  the  angular  velocity  ; 

Then,  if  the  angular  motion  is  uniform, 

(3) 


t 


If  it  is  variable,  then 


d0 

dt 


W 


We  also  have. 


ds  =  vdt  =  4/^^(92  ^.  ch^ ; 

j/n      d^  —  dr^      v^df  —  d?^         , 
.*.  dff^  = s — s ;  and 


de 

dt 


v^-©" 


(5) 


17.  Acceleration  is  the  rate  of  increase  or  decrease  of  the 
velocity.  It  is  a  velocity-increment.  The  velocity-increment 
of  an  increasing  velocity  is  considered  positive,  and  that  of  a 
decreasing  velocity,  negative. 

The  measure  of  the  acceleration,  when  it  is  uniform,  is  the 
amount  by  which  the  velocity  is  increased  (or  decreased)  in  a 


[17.]  ACCELERATION.  7 

unit  of  time.  If  the  acceleration  is  variable,  it  is  the  amount 
by  which  the  velocity  would  be  increased  in  a  unit  of  time, 
provided  the  rate  of  increase  continued  the  same  that  it  was 
lit  the  instant  considered. 


Hence,  if 

y  =  the  measure  of  the  acceleration  (or, 

,  briefly,  the 

acceleration) ; 
then,  when  the  acceleration  is  uniform, 

J,      V      dt 

and  hence,  when  it  is  variable, 

d- 
j,_  dv  _      dt  _d^s 

'^~  dt           lt~de' 

(«) 

We  also  have 

J,     dh      ds^      d's       dsi      'd^8 
^       df^df^       d6^  ^    dt^            df 

(6') 

We  thus  see  that  the  relation  between  space,  time,  and  velo- 
city are  independent  of  the  cause  which  produces  the  velocity. 

Applications  of  EQrA.TTON  (6). 

1.  Suppose  that  the  acceleration  is  constant. 
Then  in  (6)  y  will  be  constant,  and  dt  being  the  equicrescent 
variable,  we  have 

or/.  =  1+0..  (7) 

ds 
But  for  t  =  Oy-j-  =  Vq  =  the  initial  velocity  .*.  6^  =  —  i^  ; 

and  (7)  becomes 

ds  ^ftdt  +  v^t. 
Integrating  again  gives 


8  EXAMPLES.  [17.] 

But  s  =  Sofort  =  0  .'.  €2  =  8q\ 

hence  the  final  equation  is 

s=^if^  +  v,t-\-8,',  (8) 

which  gives  the  relation  between  the  space  and  time. 

Again,  multiplying  both   members  of  equation  (6)  by  ds, 
we  have 


^,fds<Ps=ffds; 


or  -^  =  2/6'  +  ro^  (9) 


di- 


.\dt=    /== 


ds 


Vv^'  + 


and  integrating,  gives 


t  ^  ^+^  +  a  (10) 

«/ 

Equation  (7)  gives  the  relation  between  the  velocity  and 

time,  and  equation  (9)  between  the  velocity  and  space. 

If  Vq  and  ^0  ^^^  ^<^th  zero,  the  preceding  equations  become 

2^ 

V  ^ft  =  ^2/^  =  — .  (11) 

*  =  i/^  =  |.=  4^^.  (12) 

tr^j^Slj^-,  (13) 

We  shall  find  hereafter  that  these  formulas  are  applicable  to 
all  cases  in  which  \hQ  force  is  constant  and  uniform. 

2.  Find  the  relation  between  the  space  and  time  when  the 
acceleration  is  naught. 
We  have 

di?-^' 
Multiply  by  dt^  integrate  twice,  and  ^e  have 

in  which  s^  and  v^  are  initial  values;  that  is,  the  body  will  have 


[17.]  EXAMPLES.  9 

passed  over  a  space  ^o  before  t  is  computed.  Vq  is  not  only  the 
initial  but  the  constant  uniform  velocity.  If  Sq  =  0,  then  s  =^ 
Vof- 

3.  If  the  acceleration  varies  directly  as  the  time  from  a  state 
of  rest,  required  the  velocity  and  space  at  the  end  of  the 
time  t. 

ILeref=  at. 

4.  Determine  the  velocity  when  the  acceleration  varies  in- 
versely as  the  distance  from  the  origin  and  is  negative ;  ory  =- 

a 
s' 

5.  Determine  the  relation  between  the  space  and  time  when 
the  acceleration  is  negative  and  varies  directly  as  the  distance 
from  the  origin  ;  or  f=  —  bs. 

Equation  (6)  becomes 

d^  =  ^^'' 
Multiplying  both  members  by  da,  we  have 
ds(p8 


df 
Integrating  gives 


=  —  ^5  ds* 


But  v=  0  ior  8  =  So  .\  C\  =^5^;  and 


^  =  b{3,'-s>)  =  v',  (14) 

ds 


or  5*  dt  = 


[ntegrating  again  gives 

5*^  =  sin-i  -  +  Co. 


Sii 


But  ^  =  0  f or  5  =  ^0  •*•  ^2  =  —  Wi 

.-.  «  =  «o  sin  (tl^  -f-  iTT).  (15) 


10 


ACCELERATION. 


[17.1 


«  5  =  0,  25  =  i-TrJ-*,  fTT^-*,  f7r5-».  iTrh-^,  fTrJ-*,  etc. ; 


SttJ-* 


OTrb  ~  *. 


"  5  =  —  5o,  ^  =      ttJ-*, 

This  is  an  example  of  periodic  motion,  of  which  we  shall 
have  examples  hereafter. 

6.  Determine  the  space  when  the  acceleration  diminishes  as 
the  square  of  the  velocity. 


When  the  acceleration  is  constant,  the  relation  between  the 
time,  space,  and  velocity  may  be  shown  by  a  triangle,  as  in 
Fig.  3.  Let  AjB  represent  the  time,  say 
four  seconds.  Divide  it  into  four  equal 
spaces,  and  each  space  will  represent  a 
second.  Draw  horizontal  lines  through  the 
points  of  division  and  limit  them  by  the 
inclined  line  AG.  The  horizontal  lines 
will  represent  the  corresponding  velocities. 
Thus  t'2  =  ge  is  the  velocity  at  the  end  of 
the  time  ^2-  The  triangle  Aho-  represents 
the  space  passed  over  during  the  first  sec- 
ond, and  ABO  the  space  passed  over  during  four  seconds. 
The  lines  de^  fK  and  ^  67  represent  the  accelerations  for  each 
second,  which  in  this  case  are  equal  to  each  other,  and  equal 
to  5c,  which  is  the  velocity  at  the  end  of  the 
first  second.  Hence,  when  the  aGceleration 
is  imiform^  the  velocity  at  the  end  of  the 
first  second  equals  the  acceleration.  This  is 
also  shown  by  Eq.  (11) ;  for  if  ^  =  1,  -y  =y! 
Equations  (12)  and  (13)  may  be  deduced  di- 
re(;tly  from  the  figure. 

If  acceleration  constantly  varies,  the  case 

may  be  represented  as  in  Fig.  4.     To  find 

rlie  tvcceleration  at  the  end  of  the  first  second,  draw  a  tangent 

a^  to  the  curve  at  the  point  «,  and  drop  the  perpendicular  ad, 


then  will  de  be  the  acceleration. 


^     de       he      dv 

But  -  =  -r  =  —  =/=  the 


ad      ah      dt 
velocity-increment,  which  is  the  same  as  Equation  (6). 


[18. 


RESOLVED  VELOCITIES  AND  ACCELERATIONS. 


11 


18.  Kesolved  velocities  and  accelerations.  When  the 
motion  is  along  a  known  path  and  at  a  known  rate,  the  projec- 
tions of  the  velocities  and  accelerations  upon  other  paths  which 
are  inclined  to  the  given  one  will  equal  the  product  of  these 
quantities  by  the  cosine  of  the  angle  between  the  paths ;  that  is, 


v^  =  V  cos  6 


.  -t.cobO,  andy 


where  v^  and/*'  are  on  the  new  path,  and  6  the  angle  between 
the  paths. 


Examples. 

1.  If  the  velocity  v  is  constant  and  along  the  line  AjB,  which 
makes  an  angle  6  with  the  line  A  O, 

then  will  the  velocity  projected  on 
A  C  also  be  constant,  and  equal  to  v 
cos  Q ;  and  on  the  line  B  (7,  equal  to 
V  sin  e.  F^°-  5. 

2.  Let  ABC\iQ  a  parabola  whose  equation  is  ^/^  =  2^aj. 
a  body  describes  the  arc  jff67with 
such  a  varying  velocity  that  its  pro- 
jection on  BD^  a  tangent  at  j5,  is 
constant,  required  the  velocity  and 
the  acceleration  parallel  to  BE. 

From  the  equation  of  the  curve 
we  have 

dx      y ' 


From  the  conditions  of  the  problem  we  have 
dy 


dt 


=  constant  =  v' ; 


but 


dt      dt  dy  jp  P 


If 


-i>- 


which  is  the  velocity  parallel  to  a? ; 


12  EXAMPLES.  ri8.| 

d}x  _v'  dy  __v'^  ^ 

hence  the  acceleration  parallel  to  the  axis  of  x  will  be  constant. 
Let  ds  be  an  element  of  the  arc,  then  will  the  velocity  along 
the  arc  be 

ds       Id^      dy^\\         ,  /         2aj\i 

^  =  5-^  =  W+^)   ="(i  +  ^)- 

3.  Determine  the  accelerations  parallel  to  the  co-ordinate  axes 
X  and  2/,  so  that  a  particle  may  describe  the  arc  of  a  parabola 
with  a  constant  velocity. 

Let  the  equation  of  the  parabola  be 

'i^='2jpx\ 

,  ^^£_ 
' '  dx      y' 

The  conditions  of  the  problem  give 

ds 

-J-  =  constant  =  v. 
dt 


r^  .  ^s  _dy    ds  ^dy  V  d^  -^  dy^  __  dy  J        dx' 
^^'dt  '~dt"dy~  dt       ~d^         ~dt^^  "^^ 

dy  __         jpv  * 

''  dt~  s/  f  j^f  ' 
and  differentiating,  gives 

d^y  _  jpvy      dy 

W^  '^  (^2  +  2/2)1  'dt 

{f  +  ff 

which  being  negative  shows  that  the  acceleration  perpendicular 
to  the  axis  of  the  parabola  constantly  diminishes. 
Similarly  we  find 

d^x  j^V 

df-  (^+^f 


nai 


ViXAMPLES. 


18 


4.  A  wheel  rolls  along  a  straight  line  with  a  uniform  velocity; 
compare  the  velocity  of  any  point  in  the  circumference  with 
that  of  the  centre. 


Fio.  7. 

Let  V  =  the  velocity  of  any  point  in  the  circumference, 
v'=  the  uniform  velocity  of  the  centre, 
r  =  the  radius  of  the  circle, 
X  =  the  abscissa  which  coincides  with  the  line  on  which 

it  rolls,  and 
y  =  the  ordinate  to  any  point  of  the  cycloid. 
Take  the  origin  at  A.     The  centre  of  the  circle  moves  at  the 
same  rate  as  the  successive  points  of  contact  B.     The  centre  is 
vertically  over  B.     The  abscissa  of  the  point  of  contact  cor- 

responding  to  any  ordinate  y  of  the  cycloid,  is  r  vei'sin"^  -; 


dt\  rj 


dy 
'  dt 


=  V 


r         dy^ 
V  2ry  —  7^  '^^  ' 

,  V2ry  —  f 


The  equation  of  the  cycloid  is 


r  versin    ~  —  (2?'y  —  't^)^ ; 


and  from  the  theorv  of  curves 


ds^  =  d^  +  df  ,'. 


ds 


v/ 


I  + 


da^ 
dy'' 


or. 


and. 


dy 

dy~^  ~2r  ^^ ' 

^^_ds^_dyd^_/Ty 
^^dt-dt    dy-y    r     ^' 


U  GRAVITT.  [19.] 

If  ^  =  0,^^=0; 

y  =  2r,  V  =  2v' ; 
yz=ir,v  =  v\ 

Hence,  at  the  instant  that  any  point  of  the  wheel  is  in  con* 
tact  with  the  straight  line,  it  has  no  velocity,  and  the  velocity 
at  the  highest  point  is  twice  that  of  the  centre. 

The  velocity  at  any  point  of  the  cycloid  is  the  same  as  if  the 
wheel  revolved  about  the  point  of  contact,  and  with  the  same 
angular  velocity  as  that  of  the  generating  circle. 

For,  the  length  of  the  chord  which  corresponds  to  the  ordi- 
nate y  is  ^/2ry^  and  hence,  if 

v:2v'  ::  V2ry  :  2r  ; 


we  have  v  =\/  -^  v\  sls  before  found. 

19.  Gravitation  is  that  natural  force  which  mutually  draws 
two  bodies  towards  each  other.  It  is  supposed  to  extend  to 
every  particle  throughout  the  universe  according  to  fixed  laws. 
The  force  of  gravity  above  the  surface  of  the  earth  diminishes 
as  the  square  of  the  distance  from  the  centre  increases,  but 
within  the  surface  it  varies  directly  as  the  distance  from  the 
centre.  If  a  body  were  elevated  one  mile  above  the  surface 
of  the  earth  it  would  lose  nearly  -^-o  of  its  weight,  which 
is  BO  small  a  quantity  that  we  may  consider  the  force  of 
gravity  for  small  elevations  above  the  surface  of  the  earth  as 
practically  constant.  But  it  is  variable  for  different  points  on 
the  surface,  being  least  at  the  equator,  and  gradually  increasing 
as  the  latitude  increases,  according  to  a  law  which  is  approxi- 
mately expressed  by  the  formula 

g  =  32.1726  -  0.08238  cos  2Z, 

in  which  Z  =  the  latitude  of  the  place, 

g  =  the  acceleration  due  to  gravity  at  the  latitude  Z, 
or  simply  the  force  of  gravity,  and 
32.1726  ft.  =  the  force  of  gravity  at  latitude  45  degrees. 


[19.]  EXAMPLES.  15 

From  this  we  find  that 

at  the  equator  g  =  g^  =  32.09022  feet,  and 
at  the  poles      g  =  gy^  =  32.25498  feet. 

The  varying  force  of  gravity  is  determined  by  means  of  a 
pendulum,  as  will  be  shown  hereafter.  It  is  impossible  to  de- 
termine the  exact  law  of  relation  between  the  force  of  gravity 
at  different  points  on  the  surface  of  the  earth,  for  it  is  not 
homogeneous  nor  an  exact  ellipsoid  of  revolution.  The  delicate 
observations  made  with  the  pendulum  show  that  any  assumed 
formula  is  subject  to  a  small  error.  (See  Mecanique  Celeste, 
and  Puissanfs  Geodesie.) 

Substituting  g  for/* in  equations  (11),  (12),  and  (13),  we  have 
the  following  equations,  which  are  applicable  to  bodies  falling 
freely  in  vacuo ; — 

2^ 

v  =  gt  =  i^2g8  =  j; 

s  =  h^=^^  =  ivt;     >  (16) 


_  -y  _     /2s  _  2s 


Examples. 

1.  A  body  falls  through  a  height  of  200  feet ;  required  the  time  of  descent 
»nd  the  acquired  velocity.     Let  g  =  32^  feet. 

Ans.  t  =  S.5S  seconds. 
V  =  113.31  feet. 

2.  A  body  is  projected  upward  with  a  velocity  of  1000  feet  per  second; 
required  the  height  of  ascent  when  it  is  brought  to  rest  by  the  force  of 
gfravity. 

Ans.  8  =  15,544  feet,  nearly. 

8.  A  body  is  dropped  into  a  well  and  four  seconds  afterwards  it  is  heard  to 
strike  the  bottom.  Required  the  depth,  the  velocity  of  sound  being  1130  feet 
per  second. 

Ans.  231  feet. 

4.  A  body  is  projected  upward  with  a  velocity  of  100  feet  per  second,  and 
at  the  same  instant  another  body  is  let  fall  from  a  height  400  feet  above  the 
other  body ;  at  what  point  will  they  meet  ? 


16  DYNAMIC  MEASURE  OF  A  FORCE.        [20,  31.  | 

5.  With  what  velocity  must  a  body  be  projected  downward  that  it  may  in  i 
seconds  overtake  another  body  which  has  already  fallen  through  a  feet  ? 

a         

Ans.  v  =  j  +  ^2ag. 

6.  Required  the  space  passed  over  by  a  falling  body  during  the  n^^  second. 

20.  Mass  •is  QUANiTrY  of  matter.  If  we  conceive  that  a 
quantity  of  matter,  say  a  cubic  foot  of  water,  earth,  stone,  or 
other  substance,  is  transported  from  place  to  place,  without 
expansion  or  contraction,  the  quantity  will  remain  the  same, 
while  its  weight  may  constantly  vary.  If  placed  at  the  centre 
of  the  earth  it  will  weigh  nothing ;  if  on  the  moon  it  will  weigh 
less  than  on  the  earth,  if  on  the  sun  it  will  weigh  more; 
and  if  at  any  place  in  the  universe  its  weight  will  be  directly 
as  the  attractive  force  of  gravity,  and  since  the  acceleration  i: 
also  directly  as  the  force  of  gravity,  we  have 


W 


constant 


for  the  same  mass  at  all  places.  This  ratio  for  any  contem- 
poraneous values  of  TFand  g  may  be  taken  as  the  measure  of 
the  mass,  as  will  be  shown  in  the  two  following  articles.  The 
weight  in  these  cases  must  be  determined  by  a  spring  balance 
or  its  equivalent. 

21.  Dynamic  measure  of  a  force.  Conceive  that  a  body  is 
perfectly  free  to  move  in  the  direction  of  the  applied  force, 
and  that  a  constant  uniform  force,  which  acts  either  as  a  pull 
or  push,  is  applied  to  the  body.  It  will  at  the  end  of  one 
second  produce  a  certain  velocity,  which  call  v^iy  If  now 
forces  of  different  intensities  be  applied  to  the  same  body  they 
will  produce  velocities  in  the  same  time  which  are  proportional 
to  the  forces ;  or 

foe  v^t^, 

in  which  f  is  the  applied  force. 

Again,  if  the  same  forces  are  applied  to  bodies  having  liiffer- 
ent  masses,  producing  the  same  velocities  in  one  second,  then 
will  the  forces  vary  directly  as  the  masses,  or, 

f  oc  M, 


f21.]  DYNAMIC  MEASURE  OP  A  FORCE.  17 

Hence,  generally,  if  uniform^  constant  forces  are  applied  tc 
different  masses  producing  velocities  v^i)  in  one  second,  then 

f  oc  Mv^^) ; 

or,  in  tlie  form  of  an  equation, 

i  =  eMv^,^;  (17) 

where  c  is  a  constant  to  be  determined. 

If  the  forces  are  constantly  varying ^  the  velocities  generated 
at  the  end  of  one  second  will  not  measure  the  intensities  at  any 
instant,  but  according  to'the  above  reasoning,  the  rate  of  varia- 
tion of  the  velocity  will  be  one  of  the  elements  of  the  measure 
of  the  force.     Hence  if 

F  =  2i  variable  force  ; 

M  =  the  mass  moved  ; 

-^  =f=  the  rate  of  variation  of  the  velocity;    or 
velocity-increment ; 
and,  -^  be  substituted  for  Vj^  in  equation  (17),  reducing 
by  equation  (6),  we  have 

F^oMf=cM%  =  cM%  (18) 

From  this  we  have 

F 
cM=  -^\ 

hence  the  value  of  cM  is  expressed  in  terms  of  the  constant 
ratio  of  the  force  F  to  that  of  the  acceler- 
ation yi 

To  determine  this  ratio  experimentally  I 
suspended  a  weight,  FT,  by  a  very  long  line 
wire.  The  wire  should  be  long,  so  that  the 
body  will  move  practically  in  a  straight 
line  for  any  arc  through  which  it  will  be 
made  to  move,  and  it  should  be  very  small, 
so  that  it  ^^^ll  contain  but  little  mass.  By 
means  of  suitable  mechanism  I  caused  a  ^iq.%. 

constant  force,  F^  to  be  applied  horizontally 
to  the  body,  thus  causing  it  to  move  sidewise,  and  determined 


t 


18  UNIT  OF  MASS.  [22.J 

the  space  over  which  it  passed  during  the  first  second.  Thia 
equalled  one-half  the  acceleration  (see  the  first  of  equations  (12) 
when  t  =  l).  I  found  when  F=  r^jyW,  that /=  1.6  feet, 
nearly  ;  and  for  I^=  j\  ^,  f=  3.2  feet,  nearly ;  and  similarly 
for  other  forces ;  lience 

cM  =  -^j  W,  nearly. 

But  the  ratio  of  F'  to  /  is  determined  most  accurately  and 
conveniently  hj  means  of  falling  bodies ;  for  f=g  =  the 
acceleration  due  to  the  force  of  gravity,  and  W  the  weight 
of  tlie  body  (which  is  a  measure  of  the  statical  effect  of  the 
force  of  gravity  upon  the  body),  hence 

W 
cM=-;  (19) 

in  which  the  values  of  W  and  ^  must  be  determined  at  the 
same  place;  but  that  place  may  be  anywhere  in  the  universe. 
The  value  of  o  is  assumed,  or  the  relation  between  c  and  31 
fixed  arbitrarily.' 


If  c  =  1 ,  we  have 


W 

Jf=y;  (20) 


and  this  is  the  expression  for  the  mass,  which  is  nearly  if  not 
quite  universally  adopted.     This  in  (18)  gives 

_      ^^d^s       W(Ps 

^=^5^  =  7  s?^  ^'^^ 

and  hence  the  dynamic  measure  of  the  pressube  which  moves 
A  BODY  is  the  product  of  the  inass  into  the  acceleration.  Thia 
is  sometimes  called  an  accelerating  force. 

If  there  are  retarding  forces,  such  as  friction,  resistance  of  the 
air  or  water,  or  forces  pulling  in  the  opposite  direction  ;  then  the 
first  member  F,  is  the  measure  of  the  unbalanced  forces  in 
pounds,  and  the  second  member  is  its  dynamic  equivalent. 

22.  Unit  of  Mass.  If  it  is  assumed  that  c  =  1,  as  in  the 
preceding  article,  the  unit  of  mass  is  virtually  fixed.  In  (20) 
if  TF=  1  and  ^  =  1,  then  M^=\\  that  is,  a  unit  of  mass 
is  the  quantity  of  matter  which  will  weigh  one  pound  at  that 


133.]  UNIT  OF  MASS.  19 

place  in  the  universe  where  the  acceleration  due  to  gravity  is 
one.  If  a  quantity  of  matter  weighs  32^  lbs.  at  a  place  where 
g  =  32|-  feet,  we  have 

hence  on  the  surface  of  the  earth  a  body  which  weighs  32J 
pounds  (nearly)  is  a  unit  of  mass. 

It  would  be  an  exact  unit  if  the  acceleration  were  exactly 
32J  feet. 

In  order  to  illustrate  this  subject  further,  suppose  that  we 
make  the  unit  of  mass  that  of  a  standard  j^ound.  Then  equa- 
tion (19)  becomes 

in  which  g^  is  the  value  of  g  at  the  latitude  of  45  degrees.  This 
value  resubstitated  in  the  same  equation  gives 

g 

and  these  values  in  equation  (18)  give 

the  final  value  of  which  is  the  same  as  (21). 

Again,  if  the  unit  of  mass  were  the  weight  of  one  cubic  foot 
of  distilled  water  at  the  place  where  g^  =  32.1801  feet,  at 
which  place  we  would  have  W=  62.3791,  and  (19)  would  give 

_  62.3791 
^       "  32.1801' 

and  this  in  the  same  equation  gives 

32.1801     W 
■^  ~  62.3791  *  g  ' 

and  these  values  in  (18)  give 

If  = -Ts ,  as  before. 

g  df' 


20  DENSITY.  I28.i 

23.  Density  is  the  mass  of  a  unit  of  volume, 
li  M  =  the  mass  of  a  body ; 
V  =  the  vohime  ;  and 
D  =  the  density ; 

then  if  the  density  is  uniform,  we  have 

If  the  density  is  variable,  let 

S  =  the  density  of  any  element,  then 
.      dM 

.•,M=fhdV  (22) 

from  which  the  mass  may  be  determined  when  8  is  a  known 
function  of  Y. 

Examples. 

1.  In  a  prismatic  bar,  if  the  density  increases  uniformly  from 
one  end  to  the  other,  being  zero  at  one  end  and  5  at  the  other, 
required  the  total  mass. 

Let  I  =  the  length  of  the  bar ; 

A  =  the  area  of  the  transverse  section  ;  and 
X  =  the  distance  from  the  zero  end  ; 

then  will 

y  =  the  density  at  a  unit's  distance  from  the  zero  end  *, 

jx  =  the  density  at  a  distance  a? ;  and 


dV=^.Adx; 


*^  6asdx      5 


2.  In  a  circular  disc  of  uniform  thickness,  if  the  density  at  a 
unit's  distance  from  the  centre  is  2,  and  increases  directly  as 
the  distance  from  the  centre,  required  the  mass  when  the  radius 
is  10. 


lai]  APPLICATIONS  OF  EQUATION  (21).  21 

3.  In  the  preceding  problem  suppose  that  the  density  in- 
creases as  the  square  of  the  radius,  required  the  mass. 

4.  In  the  preceding  problem  if  the  density  is  two  pounds  per 
cubic  foot,  required  the  weight  of  the  disc. 

6.  If  in  a  cone,  the  density  diminishes  as  the  cube  of  the 
distance  from  the  apex,  and  is  one  at  a  distance  one  from  the 
apex,  required  the  mass  of  the  cone. 

Having  established  a  unit  of  density,  we  might  properly  saj 
that  mass  is  a  certain  number  of  densities, 

24.  Applications  of  Equation  (21). 

[Obs. — If,  for  any  cause,  it  is  considered  desirable  to  omit  any  of  the  matter 
of  the  following  article,  the  author  urges  the  student  to  at  least  establish  the 
equations  for  the  acceleration  for  each  of  the  31  examples  here  given.  This 
part  belongs  purely  to  mechanics.  The  reduction  of  the  equations  belongs  to 
mathematics.  It  would  be  a  good  exercise  to  establish  the  fundamental  equa- 
tions for  all  these  examples,  before  making  any  reductions.  Such  a  course 
serves  to  impress  the  student  with  the  distinction  between  mechanical  and 
mathematical  principles.] 

\st.  If  a  body  whose  weight  is  50  pounds  is  moved  horizon- 
tally  by  a  constant  force   of  10 
pounds,  reqxdred  the  velocity  ao-  ^^^. 

quired  at  the  end  of  10  seconds  ^^m  >^^^ibx 

and  the  space  passed  over  during  ^^^  ^ 

that  time,  there  being  no  friction 
nor  other  external  resistance,  and  the  body  starting  from  rest 

Here* 


• 

ir      ^     ^^    IK         A 

F=  10  lbs. ; 

hence  (21)  gives 

d^s       F      193 

df~  M~~  30- 

Multiply  by  dt  and  integrate,  and 

^_ 

dt~ 

The  second  intecrral  is 


^  193         ,^       ^ 

-  =  ^=  — ^  +  (^,  =  0). 


22  ACCELKEATING  FORCES.  [84.] 

and  hence  for  ^  =  10  seconds,  we  have 

V  =    64.33  +  feet. 
8  =z  321.66  +  feet. 

2d.  Siij>pose  the  data  to  he  the  same  as  in  the  jpi^eceding 
example,  and  also  that  the  friction  hetween  the  body  and  the 
flane  is  5  jpounds.  Required  the  space  passed  over  in  10 
seconds. 

Here  F=  (10  —  5)  pounds. 

dJ^s  _  F       193 
'''df~  M~  60' 

3<^.  Suppose  that  a  hody  whose  weight  is  ^0  pounds  is  moved 
horizontally  hy  a  weight  of  10  lbs.,  which  is  attached  to  an  inex- 
tensihle,  but  perfectly  flexible  string  which  passes  over  a  wheel 
and  is  attached  at  the  other  end  to  the  body.  Required  the 
distance  passed  over  in  10  seconds,  if  the  string  is  without 
weight,  and  no  resistance  is  offered  by  the  wheel,  plane,  or 
string. 

soils. 


\z^ 


k' 


tolls. 

ill  mm 

Fio.  10. 

In  this  case  gravity  exerts  a  force  of  10  pounds  to  move  the 
mass,  or  i^=  10  lbs.,  and  the  mass  moved  is  that  of  both  bodies, 
or  Mz=z  (50  +  10)  -T-  32f 

^_^'_193 
di^  ~'M  ""  36  • 

The  integration  is  performed  as  before. 

Ans.  s  =  268.05  feet. 

^th.  Find  the  tension  of  the  string  in  the  preceding  example. 


[24.]  MOVING  MASSES.  28 

The  tension  will  equal  that  force  which,  if  applied  directly 
to  the  body,  as  in  Ex.  1,  will  produce  the  same  acceleration  as 
in  the  preceding  example. 

Let  jP  =  10  pounds ; 
TT  =  50  pounds ; 
T  =  tension ; 

=  the  mass  in  the  former  example ;  and 

W 

—  =  the  mass  moved  by  the  tension. 

Sence,  from  Equation  (21), 

f=  F ;  and 

tf 
W 

Eliminate yj  and  we  find 

WP 


T 


.'.  r=  8.33  lbs. 

What  must  be  the  value  of  P  so  that  the  tension  will  be  a 
maximum  or  a  minimum,  P  -h   IF  being  constant? 

6tA.  In  example  3,  what  must  he  the  weight  of  P  so  that 
the  tension  shall  be  (^)    ^^art  of  P  f 

Ans.  P  =  (n-l)  W. 

6th.  If  a  body  whose  weight  is  W  falls  freely  in  a  vacuum 
by  the  force  of  gravity,  determine  the  formulas  for  the  motion. 

Here  Mg  =  W  and  the  moving  force  F=  W\ 

g  dt 
d^s 

The  integrals  of  this  equation  will  give  Equations  (16), 
when  the  initial  space  and  velocity  are  zero.  Let  the  student 
deduce  them. 


-  =  TF- 


24  PROBLEMS  OP  [24] 

1th.  Suppose  that  the  moving  pressure  {pull  or  push)  equah 
the  weight  of  the  body,  required  the  velocity  and  space. 

Here  Mg  =  W  and  J^=  W,  hence  the  circumstances  of 
motion  will  be  the  same  as  in  the  preceding  example. 

The  forces  of  nature  produce  motion  without  apparent 
pressure,  but  this  example  shows  that  their  effect  is  the  same 
as  that  produced  by  a  push  or  pull  whose  intensity  equals  the 
weight  of  the  body,  and  hence  both  are  measured  by  pounds, 
or  their  equivalent. 

Sth.  If  the  force  F  is  constant,  show  that  Ft  =  Mv  ;  also 
Fs  —  ilfv^,  a? id   ^Ff  =  Ms.      If  F  is  variable  we    have 

Mv  =fFdt. 

^th.  Suppose  that  a  piston,  devoid  of 

friction,  is  driven  by  a  constant  steam- 
pressure  through  a  portion  of  the  length 
of  a  cylinder,  at  what  point  in  the  stroke 
must  the  pressure  be  instantly  reversed 
Fig.  11.  **^  ^0  that  the  full  stroke  shall  equal  the 
length  of  the  cylinder,  the  cylinder  being  horizontal  f 

At  the  middle  of   the  stroke.      Whatever  velocity  is  gen 
erated  through  one-half  the  stroke  will  be  destroyed  by  the 
counter  pressure  during  the  other  half. 

lOz^A.  If  the  pressure  upon  the  piston  is  500  pounds,  weight 
of  the  piston  hO  pounds,  and  the  friction  of  the  piston  in  the 
cylinder  100  po2inds,  required  the  point  in  the  stroke  at  which 
the  pressure  must  be  reversed  that  the  stroke  may  be  12  inches. 
The    uniform  effective  pressure  for  driving   the   piston  is 
500  —  100  =  4:00   lbs.,    and   the   uniform    effective   force   for 
stopping  the  motion  is  500  +  100  =  600  pounds.     The  velocity 
generated   equals    the   velocity   destroyed,   and    the    velocity 
destroyed  equals  that  which  wonld  be  generated  in  the  same 
space  by  a  force  equal  to  the  resisting  force ;  hence  if 
F=  the  effective  moving  force  ; 
s  =  the  space  through  which  it  acts ; 
V  =  the  resultant  velocity  ; 
F^  =  the  resisting  foi'ce ;  and 
s'  =  the  space  through  which  it  acts  ; 


[34.1 


ACCELERATING  FOECES. 


25 


then,  from  tlie  expression  in  Example  8,  we  have 

Fs  =  \M^^, 
and  F'a'  =  iJ/zr^, 

/.  Fs  =  F's', 
or,  F  :  F':\s'  :  s. 

In  the  example,  F=  400  lbs.,  and  F'=  600  lbs.  Let  x  =  the 
distance  from  the  starting  point  to  the  point  where  the  pressure 
must  be  revei-sed.     Then 

600  :  400  ::  a;  :  12  -  x,  .-.  x  =  7i  inches. 

11th.  If  in  the  preceding  examjple  the  _pi8ton  moves  vertically 
up  and  down,  required  the  point  at  which  the  pressure  must 
he  instantly  reversed  so  that  the  full  stroke  shall  he  12  inches. 

The  effective  driving  pressure  upward  will  be  500  —  100  — 
50  =  350  pounds,  and  the  retarding  force  will  be  500  +  100  + 
50  =  650  pounds,  and  during  the  down-stroke  the  driving  force 
is  500  +  50  —  100  =  450  pounds,  and  the  retarding  force  is 
500  -  50  +  100  =  550  pounds. 

I'zth.  A  string  passes  over  a  wheel  and  has 
a  weight  F  attached  at  one  end,  and  W  at  the 
other.  If  there  are  no  resistances  from  the 
string  or  wheel,  and  the  string  is  devoid  of 
weight,  required  the  resulting  motion. 

Suppose  W  >  P', 
then 


F=  W-  P,  and 

Tr+  P 


Pio   12. 


M 


,  dh_F___   Wj-  P 
"  dt''-  M~   W  +  P^' 
By  integrating,  we  find 

W-P 


and. 


W-hP 
W—  P 


26 


PROBLEMS  OF 


[34] 


ISth.  Required  the  tension  of  the  string  in  the  jpreceding 
examjple. 

The  tension  equals  the  weight  P,  plus  the  force  which  will 
produce  the  acceleration 

W-P 

when  applied  to  raise  P  vertically.     The  mass  multiplied  by 
the  acceleration  is  this  moving  force,  or 
P     W-P 


hence  the  tension  is 


g  •  W-\-  P^' 


P+  w-J'p^^wp 

Similarly,  it  equals  TF  minus  the  accelerating  force,  or 


TT- 


W-P 


W  = 


2WP 


W+P  W+P' 

A  complete  solution  of  this  class  of  problems  involves  the 
mass  of  the  wheel  and  frictions,  and  will  be  considered  here- 
after. 

l^th.  A  string  passes  over  a  wheel  and  has  a  weight  P 
attached  to  one  end  and  on  the  other  side  of 
the  wheel  is  a  weight  TF,  which  slides  along 
the  string.  Required  the  friction  between 
the  weight  W  and-  the  string,  so  that  the 
weight  P  will  remain  at  rest.  Also  re- 
quired the  acceleration  of  the  weight  W, 


The  friction  =  P  ; 


Fio.  13. 


Mg-. 
and,  F 

'''  df~ 


TF; 

TF-P; 

F         W-P 


hence, 


and, 


v  = 


M  "" 

W-P 
W 


W 


9 


9% 


[241  ACCELERATING  FORCES.  27 

loth.  In  the  jpreceding  exam/ple^  if  Wwere  an  animal  whoso 

vjeiyht  is  less  than   P,  required  the  acceleration  with  which 

it  must  ascend,  so  that  P  will  remain  at  rest,    '. , 

V>/ .     > 
\^th.  If  the  weight  W  descend  along  a  rough  rojpe  with  a 

given  acceleration,  required  the  acceleration  with  which  the 

body  P  must  ascend  or  descend  on  the  opposite  rope,  so  that 

the  rope  may  remain  at  rest,  no  allowance  being  made  for 

friction  on  the  wheel,  '^.-vi.i-V^]-^!^ 

(Tlie  ascent  must  be  due  tcyiblimbing  up  on  the  cord,  or  be 
produced  by  an  equivalent  result.) 

11th.  A  particle  moves  in  a  straight  line  under  the  action 
of  a  uniform  acceleration,  and  describes  spaces  s  and  s'  in  t*^ 
and  t'^^  seconds  respectively,  determine  the  accelerating  force 
and  the  velocity  of  projection. 
Let  v^  =  the  velocity  of  projection,  and 

y  =  the  acceleration  ; 
then 

/= 


t'-t} 


,  s\n  - 1)  —  si^'itf  - 1) 

and  ^0= ^^^ ^. 

If  -=s^ -,  then  2^0  =  0. 

s       2t  —  1 

18th.  If  a  perfectly  flexible  and  perfectly  smooth  rope  is 
placed  upon  a  pin,  find  in  what  time  it  will  run  itself  off , 

If  it  is  perfectly  balanced  on  the  pin  it  will  not  move,  unless 
it  receive  an  initial  velocity.  If  it  be  unbalanced,  the  weight 
of  the  unbalanced  part  will  set  it  in  motion.  Suppose  that  it 
is  balanced  and  let 

Vq  =  the  initial  velocity, 
21  =  the  length  of  the  rope, 
w  =  the  weight  of  a  unit  of  length,  and 
t  =  the  tfme. 

Take  the  origin  of  coordinates  at  the  end  of  the  rope  at 
the  instant  that  motion  begins.  When  one  end  has  descended 
9  feet,  the  other  has  ascended  tlic  same  amount,  and  hence  the 


28 


PROBLEMS  OP 


[24.] 


unbalanced  weight  will  be  ^ws, 

2wl  -r  g ;  hence  we  have 

^  _  F_  _  "i^jm    _  g 
'de"  M~  2wl^  ~'f' 

Multiply  by  ds  and  integrate,  and  we  have 


The  mass  moved  will  b€ 


d^ 


Integrating  again,  gives 


ds 


\/U'  + 


\/l"' 


0     + 


\/^'" 


£,  '-•(' 


=  ^llog 


1  + 


V/o^  +  ^    I 


[-  ,  if  s  =  «. 


l^th.  If  a  particle  moves  towards  a  centre  of  force  whtcn 
ATTRACTS  directly  as  the  distance  from  the  force,  determine 
the  motion. 

Let  jjb  =  the  absolute  force  ;  that  is,  the  acceleration  at  a 
unit's  distance  from  the  centre  due  to  the 
force;  and 


then 


8  =  the  distance ; 


d's 

A  force  is  considered  positive  in  whatever  direction  it  acts, 
and  the  j[)lus  sign  indicates  that  its  direction  of  action  is  the 
same  as  that  of  the  positive  ordinate  from  the  origin  of  coor- 
dinates, and  the  negative  sign  action  in  the  opposite  direction. 
If  a  be  the  initial  value  of  «,  we  have  (see  Ex.  5,  Art.  17) : 

v:=y/ y,{(j?  -  6-^); 
t  z=  jjb-2  (sin- -Itt)  ; 


[241  ACCELERATING  FORCES.  29 

and  the  velocity  at  the  centre  of  the  force  is  found  by  making 
*  =  0,  for  which  we  have, 

'c  =  a  Vfi  ; 
and  t=  -  -^fi-^n,  -fi-^n,  -^H^^y  etc., 

hence,  the  time  is  independent  of  the  initial  distance. 

It  may  be  proved  that  within  a  homogeneous  sphere  the 
attractive  force  varies  directly  as  the  distance  from  the  centre. 
Hence,  if  the  earth  were  such  a  sphere,  and  a  body  were  per- 
mitted to  pass  freely  through  it,  it  would  move  with  an  accele- 
rated velocity  from  the  surface  to  the  centre,  at  which  point 
the  velocity  would  be  a  maximum,  and  it  would  move  on  with 
a  retarded  velocity  and  be  brought  to  rest  at  the  surface  on  the 
opposite  side.  It  would  then  return  to  its  original  position, 
and  thus  move  to  and  fro,  like  the  oscillations  of  a  pendulum. 

The  acceleration  due  to  gravity  at  the  surface  of  the  earth 
being  ^,  and  ?•  being  the  radius,  the  absolute  force  is 

r 

and  the  time  of  passing  from  surface  to  surface  on  the  equator 
would  be 

^  A       oiAia     720,923,161        .„      .^ 

t  =  n^-  =  3.1416  V-i^j^^  =  42m.  1.6  sec. 

The  exact  dimensions  of  the  earth  are  unknown.  The  semi-polar  axis  ol 
the  earth  is,  as  determined  by 

Bessel 20,853,663  ft. 

Airy 20,853,810ft. 

Clarke     20,853,429  ft. 

The  equatorial  radius  is  not  constant,  on  account  of  the  elevations  and 
depressions  of  the  surface.  There  are  some  indications  that  the  general  form 
of  the  equator  is  an  ellipse.  Among  the  more  recent  determinations  are 
those  by  Air.  Clarke,  of  England  (1873),  and  his  result  given  below  is  considered 
by  him  as  the  most  probable  mean.     The  equatorial  radius,  is  according  to 

Beese:  20,923,596  ft. 

Airy     20,923,713  ft. 

Clarke 20,923,161ft. 


MliMii 


80  PROBLEMS  OP  [2L] 

The  determination  of  the  force  of  gravity  at  any  place  is  subject  to  small 
errors,  and  when  it  is  computed  for  different  places  the  result  may  differ  from 
the  actual  value  by  a  perceptible  amount. 

The  force  of  gravity  at  any  particular  place  is  assumed  to  be  constant,  but 
all  we  can  assert  is  that  if  it  is  variable  the  most  delicate  observations 
have  failed  to  detect  it.  But  it  is  well  known  that  the  surface  of  the  earth 
is  constantly  undergoing  changes,  being  elevated  in  some  places  and  depressed 
in  others,  and  hence,  assuming  the  law  of  gravitation  to  be  exact  and  imiversal, 
we  cannot  escape  the  conclusion  that  the  force  of  gravity  at  every  place  on  its 
surface  changes,  and  although  the  change  is  exceedingly  slight,  and  the  total 
change  may  extend  over  long  periods  of  time,  it  may  yet  be  possible,  with 
apparatus  vastly  more  delicate  than  that  now  used,  to  measure  this  change. 
It  seems  no  more  improbable  than  the  solution  of  many  problems  already 
attained — such  for  instance,  as  determining  the  relative  velocities  of  the  earth 
and  stars  by  means  of  the  spectroscope. 

'2,0th.  Supj[>ose  that  a  coiled  spring  whose  natural  length  it 
A  B^is  compressed  to  B  C,     If  one  end  rests  against  an  im- 
movable   body  B,  and  th^ 

which  is  jperfectly  free  to 
-_     .  A     7nove  horizontally,  what  will 
Fio.  14.  be   the   timd    of  movement 

from  O  to  A,  and  what  will  be  the  velocity  at  A  f 

It  is  found  by  experiment  that  the  resistance  of  a  spring  to 
compression  varies  directly  as  the  amount  of  compression,  hence 
the  action  of  the  spring  in  pushing  the  body,  will,  in  reference 
to  the  point  A,  be  the  same  as  an  attractive  force  which  varies 
directly  as  the  distance,  and  hence  it  is  similar  to  the  preceding 
example.  But  if  the  spring  is  not  attached  to  the  particle  the 
motion  will  not  be  periodic,  but  when  the  particle  has  reached 
the  point  A  it  will  leave  the  spring  and  proceed  with  a  uniform 
velocity.  If  the  spring  were  destitute  of  mass,  it  would  extend 
to  A,  and  become  instantly  at  rest,  but  because  of  the  mass  in 
it,  the  end  will  pass  A  and  afterwards  recoil  and  have  a  periodic 
motion.  If  the  body  be  attacJied  to  the  spring,  it  will  have  a 
periodic  motion,  and  the  solution  wnll  be  similar  to  the  one 
in  the  Author's  Resistance  of  Materials,  Article  19. 

Take  the  origin  at  A,  s  being  counted  to  the  left ;  suppose 
that  5  pounds  will  compress  the  spring  one  inch,  and  let  the 
total  compression  be  <x  =  4  inches.  Let  W  =  the  w^eight  of 
the  body  =  10  pounds. 


v 


134.]  ACCELERATING  FORCES.  81 

The  force  at  the  distance  of  one  foot  from  the  origin  being 
60  pounds,  the  force  at  8  feet  will  be  60.?  pounds. 


Hence, 

rs'=-^=-«»- 

or, 

g=-M8,; 

from  which 
and 

we 

find  that 

t-       *" 
\/l93' 

?;  =  4.6  4-  feet. 

2l8t.  Suppose  that  in  the  preceding  jprohlem  a  hody  whose 
weight  is  M'  is  at  B,  and  another  M"  at  O^  hoth  heing  perfectly 
free  to  move  horizontally^  required  the  time  of  movement  that 
the  distance  between  them  shall  he  equal  to  AB ;  and  the 
resultant  velocities  of  each. 

Take  the  oi-igin  at  any  convenient  point,  say  in  the  line  of 
the  bodies  and  at  a  distance  x'  to  the  left  of  M\  and  let  re"  be 
the  abscissa  of  M" ;  h  the  length  of  the  spring  after  being 
compressed  an  amount  a,  and  /i  the  force  in  pounds  which 
will  compress  it  the  iirst  unit ;  then  the  tension  of  the  spring 
when  the  length  is  s  will  be  pi{a  +  h  —  8)\  hence  we  have 

M'^^=^-n{a  +  h-»). 
From  the  first, 

dJ's^d^x"      dj^x\ 
W       di?        di^*' 
substituting, 

integrating, 

ds'  31'+    31''^  ^„;  0,^^ 


82  REPULSIVE  FORCES.  [24.] 

But-y  =  0  for  5=xf;  .'.  6i  =  -  fi^ M'M" ' 

and  integrating  again  gives 


./    M'  +  31"     ,  -1^  —  ^  .    r/1       f\\ 


and,  making  ^  =  ^  +  J,  we  have 


Jtt/ 


1       J/'Ji" 

Ji'  J£'  +  Jl"' 


which,  as  in  the  preceding  example,  is  independent  of  the 
amount  of  compression  of  the  spring. 

To  find  the  relation  between  the  absolute  velocities, 

Let  s'  =  the  space  passed  over  by  M',  and 
s"  =  the  space  passed  over  by  M'^; 

then  since  the  moving  force  is  the  same  for  both,  we  have 

Integrating,  gives 

22d,  Sv{ppose  that  the  force  varies  directly  as  the  distarioe 
from  the  centre  of  force  and  is  repulsive. 

Then 

d^s 

in  which  Vq  is  the  initial  velocity. 

23<^.  Suj[ypose  that  the  force  varies  inversely  as  the  square  of 
the  distance  from  the  centre  of  the  force  and  is  attractive. 

[This  is  the  law  of  universal  gravitation,  and  is  known  as  the  law  of  the 
inverse  squares.       While  it  is  rigr^dly  true,  so  far  as  we  know,  for  every 


[»4.J  ATTRACTIVE  FORCES.  33 

particle  of  matter  acting  upon  any  other  particle,  it  is  not  rigidly  true  for 
finite  bodies  acting  upon  other  bodies  at  a  finite  distance,  except  for  Jiomoge- 
neou^  fsphert  s,  or  spheres  composed  of  homogeneous  shells.  The  earth  bein^ 
ueither  homogeneous  nor  a  sphere,  it  will  not  be  exactly  true  that  it  attracts 
external  bodies  with  a  force  which  varies  inversely  as  the  square  of  tht 
distance  from  the  centre,  but  the  deviations  from  the  law  for  bodies  at  great 
distances  from  the  earth  will  not  be  perceptible.  We  assume  that  the  law 
applies  to  all  bodies  above  the  surface  of  the  earth,  the  centre  of  the 
force  being  at  the  centre  of  the  earth.] 

Let  the  problem  be  applied  to  the  attraction  of  the  earth, 
and 

r  =  the  radius  of  the  earth ; 

g  =  the  force  of  gravity  at  the  surface ; 

fi  =  the  absolute  force ;  and 

s  =  the  distance  from  the  centre ; 
then 

f*  =  ^^ ; 

Multiply  by  ds  and  integrate ;  observing  that  fors  =  a,v  =  0, 
and  we  have 


and 


df 


=  '''(]-l)  ('^) 


^    as  —  sr 
=  2/*— ::o-; 


\a  I  (as  —  s^)^ 


using  the  negative  sign,  because  t  and  s  are  inverse  functions 
of  each  other. 

The  second  member  may  be  put  in  a  convenient  form  for 
integration  by  adding  and  subtracting  i  a  to  the  numerator 
and  arranging  the  terms.     This  gives 

ia  —  s  —  ia 


Lilt 


(as  -  s')i 


ds 


a  —  2s     J  ads 

ds  — 


2(a5-«2)i  2{as-»^i  ' 


34  ATTEACTIVE  FOB0K8.  [ZL\ 

the  integral  of  which  is 

(as  —  s^)i  —  ^a  versin"^  —  +  O, 
^  ^  a 

But  when  s  =  a^  t  =  0  ,\  0=  ia^ ; 

From    the    circle  we  have  n  —  versin-^    —  =  ti  —  cos-wl  —  _  j  = 
cos-.(-(l -!»))=  cos-  (Ll). 

From  trigonometry  we  have  2  cos'  y  —  1  =  cos  2  y. 
Let  2  y=  cos  ~M 1 1,  then 

eoB  2y  =  _  —  1 ;  . •.  cos^  y  =  — ,   and  2^  =  cos-'  a/ L\  and 

a  «  y    a 

2y  =  2  cos-'  i/  ~  1  ^^  ^  — versin-'—   J. 

From  (a)  it  appears  that  for  s  =  0,v  =  oo;  hence  the  velocity 
at  the  centre  will  be  infinite  when  the  body  falls  from  a  finite 
distance. 

Jf  s  =  a  =  co  ,v  =  0.  If  a  body  falls  freely  from  an  infinite 
distance  to  the  earth,  we  have  in  equation  {a) 

«  =  00  ;  and 

s  =  r  =  the  radius  of  the  earth ; 

■■■•=(-¥)*. 

for  the  velocity  at  the  surface.      But  ^=g; 
.-.  V  =  {2gr)i, 
If  ^  =  32J  feet  and  r  =  3962  miles,  we  have 

Hence  the  maximum  velocity  with  which  a  body  can  reach 
the  earth  is  less  than  seven  miles  per  second. 


(34.1  ATTRACTIVE  FORCES.  35 

24M.  Sujp^oae  that  the  force  is  attbactive  and  varies  in 
versely  as  the  r^  jpower  of  the  distance. 

Then 

d?s  _  _ii. 
d^"      ^' 

dl  ^  ^j4_  l\ 1_  \ 

•*•  d^      n-\  \6'«-^       a"-*  / ' 

and  integrating,  gives 

According  to  the  tests  of  integrdbility  this  may  be  integrated 
when 

5     3      1,35 
''=  ....^,  5,  3>-  1,1-, or-....  etc., 

orn=  .  .  .  .  ^,  -,  -,0,2,or- etc. 

25  ^A.  Let  the  force  vary  inversely  as  the  square  root  of  the 
distance  and  he  attractive.  (This  is  one  of  the  special  cases 
of  the  preceding  example.) 

We  have 

d^s  _  _\i 

de"    ii' 

or,  'itt^dt^.     ~^ 


(a*  -  «♦)* 


The  negative  sign  is  taken  because  i  and  «  are  inverse  functions  of  each 
other. 

Add  and  subtract   — —    .  and  we  have 

3  vT\/a*-«* 

[-y/T  2Va 2Va  ")  ^ 


36  ATTRACTIVE   FORCES.  [24.] 

L    ^^/ «  V  ^*  —  **     ^ V «  V  ^*  —  ** J 

26ifA.  Swppose  that  the  force  is  attractive  cmd  vaHes  i'lh 
versely  as  the  distance. 

Hence 

d^s__       fji 

m  which  s  =  a  for  v  =  0.      Hence  the  time  from  s  =a  to 
«  =  0,  is 

._     1       ro    ds        _    /ir\i 

t  —  '^ 


Let  I  log  —  i  =  y  ;  then  f or  «  =  a,  y  =  0  and  for  «  =  0,  y  =  oo .    Squaring 
and  passing  to  exponentials,  we  have 

log  —  =  2^  . •.—  =  6^   ,    or  «  =  a  «"*  ; 
.•.  d«  =  —  ae  ^  .%y  dy; 

■00 


.'.  f 


"(^)V  ^^''^  =  "(2^)* 


This  is  called   a  gamma-function^  and  a  method  of  integrating  it  is  aa 
follows : — 

Since  functions  of  the  same  form  integrated  between  the  same  limits  are 
independent  of  the  variables  and  have  the  same  value,  therefore 


f241  ATTRACTIVE  FORCES.  87 

Knd    I       e'^dy    j     ^^  at  =     I       /     e-^  dy    I  . 

Also  the  left  hand  member  will  be  of  the  same  value  if  the  sign  of  integra- 
tion be  placed  over  the  whole  of  it,  since  the  actual  integration  will  be 
performed  in  the  same  order^^  hence 

r—      /^oo  —1  2         /*co      .-KO 

I    e-^dy    \     =  /        e-^-^dydt 

^/0        Jo 

in  which  jr  =  fu ;  . '.  dy  =  t  du.     Integrating  in  reference  to  f ,  we  have 

-<2  (1  +  w2) 


dUj 


3  (1  +  t^2) 

du 
which  for  ^  =  oo  becomes  zero,  and  for  t  =  0  becomes r-  ,  and  the  in- 

tegral  of  this  is  i  taa'^u^  which  is  zero  for  w  =  0,  and  i  tt  f or  «  =  oo  ; 


■Jo    ''^'^=\s/~- 


(See  also  M^c.  Celeste,  p.  151  [1534  O].) 
Or  we  may  proceed  as  follows  : — 

Let  g-t^  =  x.'.dt=-{^  log  x)~^^', 

.-.  /      e-^^dt=  f  -ii-logxy-idx. 
Jo  J\ 

Let  X  =z  a^  and  consider  a  less  than  unity  ;  then  log  a  will  be  negative, 
and  log  x  =  y^i—  log  a) ; 

,\  dx  =  dy  2y  dy  {— log  a) ; 


38  ATTRACTIVE   FORCES.  [24] 

which  substituted  above  gives 

J^    e-t  dt  =J   -  i  ( -  log  xT^da;  =  (  -  log  ay  J     aP  dy. 
Dividing  by  (  —  log  ay  and  multiplying  both  sides  by  —  i  da,  we  have 
J    -:^{- log a)~^ da  J    -i  {-log x)'^dx  =J      J    -laV^dyda. 
Integrating  the  second  member  first  in  regard  to  a,  gives 


-\ 


y2  +  l 


which  between  the  limits  of  0  and  1  gives  \ ;  the  integral  o(  which  if 

1  -+-^2 

\  tan-iy  which  between  the  limits  of  oo  and  0  gives  lir. 

.'.    I  -i{- logoff  da  J    -i{-\ogx)~^dx 

=      f  -i{-\ogx)-hx         =i7r; 


•*•     /       e~^  dt  =  iVir. 

(See  Mec.  CSleste,  Vol.  iv.  p.  487,  Nos.  [8319]  to  [8331].  Chauvenet's 
Spherical  Astronomy^  VoL  i.  p.  153.  Tod^water' b  Integral  Calculus.  Price's 
Infinitesimal  Calculus.) 

Sometimes  the  integration  of  an  exponential  quantity  becomes  apparent 
by  first  difEerentiating  a  similar  one.  Thus,  to  integrate  t  e~^  dt,  first  differ- 
entiate e-i^-    We  have  d e-t^=  er^d {-  t^)  =  e-^^ {-  2tdt)  =  -  2te-^  dt ; 

.  •.  Jde-i^  =  -  2  Jte-^^dt. 

But  the  first  member  is  the  integral  of  the  differential,  and  hence  is  the 

Ji  2 

quantity  itself,  or  «  ^  ,  and  hence  the  required  integral  is  -^  e~^  . 

27^A.  Swppose  that  two  hodies  have  their  centres  at  A  and 

A'  respectively^  and 

j^_^ ^     i     \ ^j^     ATTRACT  ajparticle  at 

Fio.  16.  jp  with  forces  which 

vary  as  the  distances  from  A  and  A', 


[24.]  ATTRACTIVE  FORCES.  89 

Let  G  be  midway  between  A  and  A' ; 

A0=  CA'  =  a', 
Ob  —  s  =  any  variable  distance ; 
and  let    (a  =  fi'  z=z  the  absolute  forces  of  the  bodies  A  and  A' 
respectively. 

Then 

■^  =  [I  {a  -  s)  -  f*  {a  +  8)  =  -  2fis; 

.•.g-=2,(^-^); 

and  integrating  again  gives 

s  =  ccos  t  a/2ju. 

28th,  Suppose  that  a  particle  is  projected  with  a  velocity  u 
into  a  medium  which  resists  as  the  square  of  the  velocity  / 
determine  the  circumstances  of  motion. 

Take  the  origin  at  the  point  of  projection,  and  the  axis  8  to 
coincide  with  the  path  of  the  body. 

Let  fi  =  the  absolute  resistance — or  the  resistance  of  the 
medium  when  the  velocity  is  unity ; 

(ds\^ 
-=- 1  =  the  resistance  for  any  velocity ; 

^  _  [ds\^ 

•*•  df~^  \dt)  ' 

d\ 


"fids. 

dt 

ds 
And  integrating  between  the  initial  limits,  g  =  0  f or  -^  =  f#, 

Cut 

and  the  general  limits,  we  have 

ds 
di 


^^S:^- ^^g^  =  -f^5 


4:0  RESISTING  FORCES.  [24.  ^ 

d^ 
or, 


=  -lis; 

ds 

•*•  dt 

=  ue-f^ 

vdt 

=  ^'ds. 

or, 

Integrating  again,  observing  that  ^  =  0  for  5  =  C,  we  have 

fjlUt=^  —1. 

The  velocity  becomes  zero  only  when  s  =  oo  . 

29th.  A  heavy  hody  falls  in  the  air  hy  the  force  of  gravity, 
the  resistance  of  the  air  varying  as  the  square  of  the  velocity  ; 
determine  the  motion. 

Take  the  origin  at  the  starting  point,  and 

Let  K  =  the  resistance  of  the  body  for  a  unit  of  velocity ; 
s  =  the  distance  from  the  initial  point,  positive  down- 
wards ; 
f  —  the  time  of  falling  through  distance  s ; 

then  -^  =  0  for  t  and  5  =  0; 
dt 

-y-  =  V  for  t  ==  t  and  «  =  « ;  and 
dt 

«j  -^  j  =  the  resistance  of  the  air  at  any  point,  and  acta 
upwards ; 
and  g  =  the  accelerating  force  downward  ; 

hence,  the  resultant  acceleration  is  the  difference  of  the  two,  oi 


or, 


d^s  idsV  ,  . 

Kdt      '~  K  W» 


[24.]  FALLING  BODIES.  41 


-0 

Separating  this  into  two  partial  fractions,  and  integrating,  gives 


-=•(;) 


Passing  to  exponentials  gives 

""-dt-yKj    ^2t{Kg)^_^^'  (J) 

which  gives  the  velocity  in  terras  of  the  time.     To  find  it  in  terms  of  the 
space,  multiply  equation  (a)  by  ds  and  put  it  under  the  form 

dl- 


0 


Proceeding  as  before,  obeerring  tbe  proper  limits,  we  find 

..=_iogidl' 

.    d8  ' 


If  a  =  CO,  V 


=  a/-  ,     and  hence  the  velocity  tends  towards  a  constant. 


Prom  eqnatlor  (ft),  multiplying  the  terms  of  the  fraction  by  e~^(f5')', 
and  observing  that  the  numerator  becomes  the  diflferential  of  the  denominator 
integrating,  and  passing  to  exponentials,  we  have, 

which  gives  the  space  in  terms  of  the  time. 

A  neat  solution  of  equation  (a)  may  be  found  by  Lagrange^s  method  of 
Variation  of  Parameters. 


4:2  FALLING  BODIES.  [24.J 

SOth.  Suppose  that  the  hody  is  jpr ejected  ujpward  in  the  air^ 
having  the  same  coe-ffiGient  of  resistance  as  in  the  jpreceding 
example. 

Take  the  origin  at  the  point  of  propulsion,  u  being  the 
initial  velocity ;  then 

d^s       d  ds  (dsV 

d?'''^dtdt  =  ^^^Adt)'^  W 

hence,  vdt  =  —         ,  ,  , ., ; 

fc'^Xdt) 

••■■'=-(5)*!'-' &)*(*)--"-' (;-)'•!• 

ds 
Solving  this  equation  for  —  ;  we  have, 

Cut 

dt      \k)     y/J+  u  \/)r  tan  t  V^g' 

Substitute  sin  i  V^/c^-f- cos  «  V  Kg  tor  tan  t^Kg  and  the  numerator  be- 
comes the  differential  of  the  denominator,  and  observing  that  t=0  for  8  =  0, 
we  have 

1         u  Vk^  Bin  tV Kg  +   V g  cos  tV Kg 

.  =  -  log :^  ; 

which  gives  the  space  in  terms  of  the  time. 
Multiply  equation  («)  by  ds  and  it  may  be  put  under  the  form 

"*'  =  -■ — TC? 


',  ^  (IJ 

Integrating,  observing  that  u  is  the  initial  velocity,  and 


g  +  K 


© 


2  ie»  =  —  log  -—— — -H-; 


[24.]  IN  A  RESISTING   MEDIUM.  43 

da  ^    -2<8      g  f.       -2*3  \  ^ 

At  the  highest  point  v  =  0,  which  in  (/)  and  (g)  gives 
and.  5  =  I- log  A  +  juA. 


m 


(*•) 


Substitute  this  value  of  s  in  equation  (c)  of  the  preceding  example,  and  we 
have 


r    g  +  "1*2 

which  gives  the  velocity  in  descending  to  the  point  from  which  it  started  ; 
and  as  it  is  less  than  u,  the  velocity  of  return  will  be  less  than  that  with 
which  it  was  thrown  upward.  This  is  because  the  resistance  of  the  air  is 
against  the  velocity  during  the  entire  movement,  both  upwards  and  down- 
wards. 

The  same  value  of  «  (Eq.  (i)  )  substituted  in  (d)  of  the  preceding  example 
gives  the  time  of  descent, 

^  =  .-4^  log  ^^  +  -^  +  «  V?: 
^  V"^  Vg  +  Ku^-u  ^1^ 

which  differs  from  the  time  of  the  ascent,  as  given  by  (h)  above. 

dlst.  Suppose  that  the  force  is  attractive  and  varies  inversely 
as  the  cube  of  the  distance,  and  that  the  medium  resists  as  the 
square  of  the  velocity,  and  as  the  square  of  the  de^isity,  the 
density  varying  inversely  as  the  distance  from  the  origin. 

Let  K  =  the  coefficient  of  resistance,  being  the  resistance 
for  a  unit  of  density  of  the  medium  and 
a  unit  of  velocity 


44  WORK.  [25. 1 

then   -g  I  3t)     =  the  resistance  at  any  point. 

Multiply  by  2ds,  and  we  have  ^  "^    1  "        ^^ 

This  is  a  linear   diflEerential  equation  of  which  the  integrating  factor  ia 

6  .       The  initial  values  are  i  =  0,  and  8  =  atorv  =  u; 

—       ^  2k  — «     —       2k  —  a      — 

—  e  « -—  e  a     U 


•••'^©-'-'-Sj 


which  gives  the  velocity  in  terms  of  the  space.     The  final  integral  cannot  be 
found. 

25.  Work  and  Yis  Yiva  {or  living  force.) — ^Resuming 
equation  (21),  and  multiplying  both  members  by  ds^  we  have 

Fds  =  Mi^ds. 
df 

Integrating  between  the  limits,  v  =  v^ior  s  =  0 ;  and  v  =  t; 
for  5  =  5,  we  have 

fFds  =  iJf  (^2  -  -yo').  (23) 

If  Vp  =  0,we  have 

fFdsr=.\Mv',  (24) 

The  expression  \Mv^  is  called  the  vis  viva  (or  living  force) 
of  a  body  whose  mass  is  M  and  velocity  v.  Its  physical  im- 
portance is  determined  from  the  first  member  of  the  equation, 
which  is  called  the  work  done  by  a  force  i^in  the  space  s. 
Hence  the  vis  viva  equals  the  work  done  hy  the  moving 
force. 

Work,  mechanically^  is  overcoming  resistance.  It  requires  a 
certain  amount  of  work  to  raise  one  pound  one  foot,  and  twice 


[35.J  WORK.  45 

that  amount  to  raise  two  pounds  one  foot,  or  one  pound  two 
feet.  Similarly,  if  it  requires  100  pounds  to  move  a  load  on  a 
horizontal  plane,  a  certain  amount  of  work  will  be  accomplished 
in  moving  it  one  foot,  twice  that  amount  in  moving  it  two  feet, 
and  so  on.     Hence,  generally,  if 

i^=  a  constant  force  which  overcomes  a  constant  resist- 
ance, and 
a  =  the  space  over  which  i^acts  projected  on  the  action- 
line  of  the  force,  then 

Wor^=  U-Fs;  (25) 

and  similarly,  if 

F=:  a  variable  force,  then 
Work=  U=XFds;  (26) 

and  if  7^  is  a  function  of  s  we  have 


U=fFd8, 


The  UNIT  of  work  is  one  pound  raised  vertically  one  foot. 

The  total  work,  according  to  equation  (25),  is  independent 
of  the  time,  since  the  space  may  be  accomplished  in  a  longer 
or  shorter  time. 

But  implicitly  it  is  a  function  of  the  time  and  velocity.  If 
the  work  be  done  at  a  uniform  rate,  we  have 

8  =  Vty 


Fa  =  Fvt, 


and 

If  ^  =  1,  we  have 

I  Fb,  (28) 

which  is  called  the  Dynamic  Effect,  or  Mechanical  Poioer. 

Mechanical  Power  is  the  rate  of  doing  work.  It  is  meas- 
ured by  the  amount  of  work  done,  or  which  the  agent  is  capa- 
ble of  doing,  in  a  unit  of  time  when  working  uniformly.  The 
unit  most  commonly  employed  is  called  the  horse-jpower,  which 
equals  33,000  pounds  raised  one  foot  per  minute. 

Every  moving  body  on  the  surface  of  the  earth  does  work,  for  it  overcomes 
a  resistance,  whether  it  be  friction  or  resistance  of  the  air,  or  some  other 
resistance.     The  same  is  true  of  every  body  in  the  universe,  unless  it  moves 


46  WORK.  125.1 

in  a  non-resisting  medium.*  Animals  work  not  only  as  beasts  of  burden, 
but  in  their  sports  and  efforts  to  maintain  life  ;  water  as  it  courses  the  stream 
wears  its  baoks  or  the  bed,  or  turns  machinery  ;  wind  fills  the  sail  and  drives 
the  vessel,  or  turns  the  windmill,  or  in  the  fury  of  the  tornado  levels  the 
forest,  and  often  destroys  the  works  of  man.  The  raising  of  water  into  the 
air  by  means  of  evaporation ;  the  wearing  down  of  hills  and  mountains  by  the 
operations  of  nature;  the  destruction  which  follows  the  lightning-stroke, 
etc.,  are  examples  of  work. 

Work  may  be  useful  or  prejudicial.  That  work  is  useful 
which  is  directly  instrumental  in  producing  useful  effects,  and 
prejudicial  when  it  wears  the  machinery  which  produces  it. 
Thus  in  drawing  a  train  of  cars,  the  useful  work  is  performed 
in  moving  the  train,  but  the  prejudicial  work  is  overcoming  the 
friction  of  the  axles,  the  friction  on  the  track,  the  resistance  of 
the  air,  the  resistance  of  gravity  on  up  grades,  etc.  It  is  not 
always  possible  to  draw  a  practical  line  between  the  useful  and 
prejudicial  works,  but  the  sum  of  the  two  always  equals  the 
total  work  done,  and  hence  for  economy  the  latter  should  be 
reduced  as  much  as  possible. 

In  order  to  determine  practically  the  work  done,  the  inten- 
sity of  the  force  and  the  space  over  which  it  acts  must  be 
measured  simultaneously.  Some  form  of  spring  balance  is 
commonly  used  to  measure  the  force,  and  when  thus  employed 
is  called  a  Dynamometer.  It  is  placed  between  the  moving 
force  and  the  resistance,  and  the  reading  may  be  observed,  or 
autographically  registered  by  means  of  suitable  mechanism. 
The  corresponding  space  may  also  be  measured  directly,  or 
secured  automatically.  There  are  many  devices  for  securing 
these  ends,  and  not  a  few  make  both  records  automatically  and 
simultaneously. 

If  the  force  is  not  a  continuous  function  of  the  space,  equa- 
tion (26)  must  be  used.  The  result  may  be  shown  graphically 
by  laying  off  on  the  abscissa,  AB,  the  distances  ao,  ce,  etc., 
proportional  to  the  spaces,  and  erecting  ordinates  ab^  cd^  ef^  etc., 
proportional  to  the  corresponding  forces,  and  joining  their 
upper  ends  by  a  broken  line,  or,  what  is  better,  by  a  line  which 

*  All  space  is  filled  with  something,  since  light  is  transmitted  from  all 
directions.  But  is  it  not  possible  that  there  may  be  a  something  through 
which  bodies  may  mo^e  without  resistance  ? 


[95.] 


WOBK. 


41 


is  slightly  curved,  the  amount  and  direction  of  curvature 
being  indicated  by  the 
broken  line  previously 
constructed ;  and  the 
area  thus  inclosed  will 
represent  the  work. 
The  area  will  be  giv- 
en    by    the    formula 

Simpson's  rule  for  determining  the  area  is: — 

Divide  the  abscissa  AB  into  an  even  number  of  equal  jparts^ 
erect  ordinates  at  the  points  of  division^  and  number  them 
in  the  order  of  the  natural  numbers.  Add  together  four  times 
the  even  ordinates^  twice  the  odd  ordinates  and  the  extreme 
ordinates^  and  multiply  the  su7n  by  one  third  of  the  distance 
between  any  two  consecutive  ordinates. 

If  yi,  2^2?  2/3?  etc.,  are  the  successive  ordinates,  and  I  the  dis- 
tance between  any  two  consecutive  ones,  the  rule  is  expressed 
algebraically  as  follows  : — 

Area  =  il  {y,  +  ^y^  ^  2y, -^  ^y, +  y,).  (29) 

If  the  applied  pressure,  F]  is  exerted  against  a  body  which 
is  perfectly  free  to  move,  generating  a  velocity  v,  then  the  work 
which  has  been  expended  is,  equation  (24),  ^M^.  This  is 
called  stored  work,  and  the  amount  of  work  wliicli  will  be  done 
by  the  moving  body  in  being  brought  to  rest  will  be  the  same 
amount.  If  the  body  is  not  perfectly  free  the  quantity  ^Mv^ 
is  the  quantity  of  work  which  has  been  expended  by  so  much 
of  the  applied  force  as  exceeds  that  which  is  necessary  in 
overcoming  the  frictional  resistance.  Thus  a  locomotive  starts 
a  train  from  rest,  and  when  the  velocity  is  small  the  power 
exerted  by  the  locomotive  may  exceed  considerably  the  resist- 
ances of  friction,  air,  etc.,  and  produce  an  increasing  velocity, 
until  the  resistances  equal  constantly  the  tractive  force  of  the 
locomotive,  after  which  the  velocity  will  be  uniform.  The 
work  done  by  the  locomotive  in  producing  the  velocity  v  in 
excess  of  that  done  in  overcoming  the  resistances  will  be^Jfy*, 
in  which  Jf  is  the  mass  of  the  train,  including  the  locomotive. 


48  EXAMPLES.  [25.J 

We  see  that  double  tlie  velocity  produces  four  times  the 
work.  This  is  because  twice  the  force  produces  twice  the 
velocity,  and  hence  the  body  will  pass  over  twice  the  space  in 
the  same  time,  so  that  in  producing  double  the  velocity  we 
have  ^K^s  =  4:F8,  and  similarly  for  other  velocities. 

[We  have  no  single  word  to  express  the  unit  of  living  force.  If  a  unit  of 
mass  moving  with  a  velocity  of  one  foot  per  second  be  the  unit  of  living 
force,  and  be  called  a  Dynam^  then  would  the  living  force  for  any  velocity 
and  mass  be  a  certain  number  of  Dynams.'\ 

Since  worh  is  not  force,  but  the  effect  of  a  force  exerted 
through  a  certain  space,  independently  of  the  time,  we  call  it, 
for  the  sake  of  brevity^,  sjpme-effect. 

Vis  viva,  or  living  force,  is  not  force,  but  it  equals  the 
work  stored  in  a  moving  mass.     It  equals  the  sjpace-effect. 

[The  expression  Mi^  was  called  the  vis  viva  in  the  first  edi- 
tion of  this  work,  and  is  still  so  defined  by  many  writers  ;  but 
there  appears  to  be  a  growing  tendency  towards  the  general 
adoption  of  the  definition  given  in  the  text.  It  is  immaterial 
which  is  used,  provided  it  is  always  used  in  the  same  sense.] 

Examples. 

1.  A  body  whose  weight  is  10  pounds  is  moving  with  a 
velocity  of  25  feet  per  second ;  required  the  amount  of  work 
which  will  be  done  in  bringing  it  to  rest. 

Ans.  97.2  foot-pounds. 

2.  A  body  falls  by  the  force  of  gravity  through  a  height  of 
h  feet ;  required  the  work  stored  in  it. 

Let  W  =  the  weight  of  the  body, 
M  =  the  mass  of  the  body, 
g  =  acceleration  due  to  gravity,  and 
V  =  the  final  velocity,  then 
^  =  2gh,  QXid  Mg  =  W; 

W 


[26.]  ENERGY.  49 

3.  A  body  whose  weight  is  100  pounds  is  moving  on  a  hori- 
zontal plane  with  a  velocity  of  15  feet  per  second;  how  far 
will  it  go  before  it  is  brought  to  rest,  if  the  friction  is  con- 
stantly 10  lbs  ?   ?^*  '^^'^""^ 

Ans.  =  34.9  ■+■  ft. 

4.  A  hammer  whose  weight  is  2000  pounds  has  a  velocity 
of  20  feet  per  second  ;  how  far  will  it  drive  a  pile  if  tlie 
constant  resistance  is  10,000  pounds,  supposing  that  the  whole 
vis  viva  is  expended  in  driving  the  pile  ?   ■  ^ 

5.  If  a  train  of  cars  whose  weight  is  100,000  pounds  is 
moving  with  a  velocity  of  40  miles  per  hour,  liow  far  will  it 
move  before  it  is  brought  to  rest  by  the  force  of  friction,  the 
friction  being  8  pounds  per  ton,  or  -^-^^j  ^^  ^^^^.  ^^^al  weight  ? 

6.  If  a  train  of  cars  weighs  300  tons,  and  the  frictional  3^c<,«  ,  c 
resistance  to  its  movement  is  8  pounds  per  ton  ;  required  the  'fiT^ 
horse-power  which  is  necessary  to  overcome  this  resistance  oXif^.i  ,^ 
the  rate  of  40  miles  per  hour. 

Ans,  256. 

T.  If  the  area  of  a  steam  piston  is  75  square  inches,  and  the 
steam  pressure  is  60  pounds  per  square  inch,  and  the  velocity 
of  the  piston  is  200  feet  per  minute,  required  the  horse-power 
developed  by  the  steam.   "tj-sT.Go.voo   rz  'i^'^  o^yo  OU 

8.  If  a  stream  of  water  passes  over  a  dam  and  falls  through 
a  vertical  height  of  16  feet,  and  the  transverse  section  of  the 
stream  at  the  foot  of  the  fall  is  one  square  foot,  required  the 
hoi*se-power  that  is  constantly  developed. 

Let  g  =32^  feet,  and  the  weight  of  a  cubic  foot  of  water, 

9.  A  steam  hammer  falls  vertically  through  a  height  of  3 
feet  under  the  action  of  its  own  w^eight  and  a  steam  pressure 
of  1000  pounds.  If  the  weight  of  the  hammer  is  500  pounds, 
required  the  amount  of  work  which  it  can  do  at  the  end  of  the 
fall.     li:^     •/^-.  3    -r    .;■  '■'    '  -  -  ---'^  ';     r'^-     ['  , 

26.  Energy  is  the  capacity  of  an  agent  for  doing  work. 
The  energy  of  a  moving  body  is  called  actxial  or  Kinetic  energy^ 
aiid  is  ex})ressed  by  \Mi^,    But  bodies  not  in  motion  may  have 
4 


50  MOMENTUM.  [27.) 

a  capacity  for  work  when  the  restraining  forces  are  removed. 
Thus  a  spring  under  strain,  water  stored  in  a  mill-dam,  steam 
in  a  boiler,  bodies  supported  at  an  elevation,  etc.,  are  examples 
of  stored  work  which  is  latent.  This  is  called  Potential 
energy.  A  inoving  body  may  possess  potential  energy  entirely 
distinct  from  the  actual.  Thus,  a  locomotive  boiler  containing 
steam,  may  be  moved  on  a  track,  and  the  kinetic  energy  would 
be  expressed  by  ^J/^,  in  which  M  is  the  mass  of  the  boiler, 
but  the  potential  energy  would  be  the  amount  of  work  which 
the  steam  is  capable  of  doing  when  used  to  run  machinery,  or 
is  otherwise  employed.  These  principles  have  been  general- 
ized into  a  law  called  the  Conservation  of  energy^  which 
implies  that  the  total  energy,  including  both  Kinetic  and  Poten- 
tial, in  the  universe  remains  constant.  It  is  made  the  funda- 
mental theorem  of  modern  physical  science. 

The  energy  stored  in  a  moving  body  is  not  changed  by 
changing  the  direction  of  its  path,  provided  the  velocity  is  not 
changed ;  for  its  energy  will  be  constantly  expressed  by  ^M'i^. 
Such  a  change  may  be  secured  by  a  force  acting  continually 
normal  to  the  path  of  the  moving  body;  and  hence  we  say 
that  a  force  which  acts  continually  perpendicular  to  the  path 
of  a  moving  body  does  no  work  upon  the  hody.  Thus,  if  a 
body  is  secured  to  a  point  by  a  cord  so  that  it  is  compelled  to 
move  in  the  circumference  of  a  circle ;  the  tension  of  the 
string  does  no  work,  and  the  vis  viva  is  not  affected  by  the 
body  being  constantly  deflected  from  a  rectilinear  path. 

MOMENTUM. 

27»  Resuming  again  equation  (21),  multiplying  by  dt,  and 
integrating  gives, 

The  expression  Mv  is  called  momentum,  and  by  comparing 
it  with  the  first  member  of  the  equation  we  see  that  it  is  the 
effect  of  the  force  F  acting  during  the  time  t,  and  is  indepen- 
dent of  the  space.  For  the  sake  of  brevity  we  may  call  the 
momentum  a  time-effect. 


f27.J  MOMENTUM.  51 


/ 


If  the  body  has  an  initial  velocity  we  have 

Fdt  =  M{v-Vo);  (31) 

'to 

which  is  the  momentum  gained  or  lost  in  passing  from  a 
velocity  Vq  to  v. 

Momentum  is  sometimes  called  qicantiti/  of  motion,  ou 
account  of  its  analogy  to  some  other  quantities.  Thus  the 
intensity  of  heat  depends  upon  temperature,  and  is  measured 
in  degrees ;  but  the  quantity  of  heat  depends  upon  the  volume 
of  the  body  containing  the  heat  and  its  intensity.  The  inten- 
sity of  light  may  be  uniform  over  a  given  surface,  and  will  be 
measured  by  the  light  on  a  unit  of  surface ;  but  the  quantity 
is  the  product  of  tlie  area  multiplied  by  the  intensity.  The 
intensity  of  gravity  is  measured  by  the  acceleration  which  is 
produced  in  a  falling  body,  and  is  independent  of  the  mass  of 
the  body ;  but  the  quantity  of  gravity  (or  total  force)  is  the 
product  of  the  mass  by  the  intensity  (or  Mg).  Similarly  with 
momentum.  The  velocity  represents  the  intensity  of  the 
motion,  and  is  independent  of  the  mass  of  the  body  ;  but  the 
quantity  of  motion  is  the  product  of  the  mass  multiplied  by 
the  velocity. 

Differentiating  (30)  and  reducing,  gives 

which  is  the  same  as  (18),  and  in  which  -37  is  a  velocity-incre- 
ment ;  hence  the  momentum  invpressed  each  instant  is  a 
measure  of  the  Tnoving  force. 

If  the  force  F  is  constant  we  have  from  (30), 
Ft  =  Mv', 
and  for  another  force  F'  acting  during  the  same  time 
F't  =  M'v' ', 
.-.  F\F'  \\Mv\  M'v' ; 
lience,  the  forces  are  directly  as  the  momenta  produced  by 
them  respectively. 


52  IMPACT.  [28.] 

If  the  forces  are  variable,  let 

J    Fdt=:Q  =  Mv,  and  J   F'dt  =  §'  =  M'v' ; 

then  Q\Q'  \\Mv\  M'v' ; 

hence  the  time-effects  are  directly  as  the  momenta  impressed. 

We  thus  have  several  distinct  quantities  growing  out  of  equation  (21)  of 
which  the  English  units  are  as  follows  : — 

The  unit  of  force,  i^,  is 1  Ih. 

The  unit  of  work  or  space  effect  is 1  lb  x  1  ft. 

The  unit  of  vis  viva  is 1  tb  of  mass  x  1^  ft.  x  1  sec. 

The  unit  of  momentum 1  lb  of  mass  x  1  ft.  x  1  sec. 

IMPULSE. 

28.  An  impulse  is  the  effect  of  a  Vlow.  When  one  body 
strikes  another,  an  impact  is  said  to  take  place,  and  certain 
effects  are  produced  upon  the  bodies.  These  effects  are  pro- 
duced in  an  exceedingly  short  time,  and  for  this  reason  they 
are  sometimes  called  instantaneous  forces;  which,  being 
strictly  defined,  means  a  force  which  produces  its  effect  in- 
stantly ^  requiring  no  time  for  its  action ;  but  no  such  force 
exists  in  nature.  The  law  of  action  during  impact  is  not  gen- 
erally known,  but  it  must  be  some  function  of  the  time. 

Resuming  equation  (31),  we  have 


J  Fdt  =  M{v  -  v^) ; 


which  is  true,  whatever  be  the  relation  between  the  force  i^^and 
the  time  t    If  the  initial  velocity  of  the  body  be  zero,  we  have 

^0  =  0, 
and  f  Fdt  =  MV=  Q.  • 

We  see  from  the  above  equation  that  as  t  diminishes  F 
must  increase  to  produce  the  same  effect.     We  see  that  in  this 


[28.1  IMPULSE.  63 

case  the  first  member  is  the  time  effect  of  an  impulse,  and  the 
second  member  measures  its  effect  in  producing  a  change  of 
velocity.     Calling  this  value  Q,  we  have 

Q  =  M{v-  V,)  =  MV,  (31a) 

Hence^  the  measure  of  an  impulse  in  jproducing  a  change  of 
velocity  of  a  body  is  the  increased  {or  decreased)  7nom,entum, 
produced  in  the  body. 

This  is  the  same  as  when  the  force  and  time  are  finite.  If 
the  force  were  strictly  instantaneous  the  velocity  would  be 
changed  from  v©  to  v  without  moving  the  body,  since  it  would 
have  no  time  in  which  to  move  it. 

Similarly  from  equation  (23)  we  have 

f  Fds^\M{v'-vi')\ 

in  which  for  an  impulse  jT  will  be  indefinitely  large;  and 
hence  the  work  done  hy  an  impidse  is  measured  in  the  sams 
way  as  for  finite  forces. 

All  the  effects  therefore  of  an  impulse  are  measured  in  the 
same  way  as  the  total  effects  produced  by  a  finite  force. 

In  regard  to  forces^  we  investigate  their  laws  of  action  ;  or 
having  those  laws  and  the  initial  condition  of  the  body  we 
may  determine  the  velocity,  energy,  or  position  of  the  body  at 
any  instant  of  time  or  at  any  point  in  space,  and  hence  we  may 
determine  final  results ;  but  in  regard  to  impulses  we  deter- 
mine only  certain  final  results  without  assuming  to  know  any- 
thing of  the  laws  of  action  of  the  forces,  or  of  the  time  or 
space  occupied  in  producing  the  effect. 

The    terms  ^''Impulsive  force, "  and  ^'Instantaneous  force, 
are  frequently  used  to  denote  the  effect  of  an    Impact ;      but 
since  the  effect  is  not  a  force,  they  are  ambiguous,  and  the 
terra   Impulse  appears  to  be  more  appropriate. 

An  incessant  force  may  be  considered  as  the  action  of  an 
infinite  number  of  infinitesimal  impulses  in  a  finite  time. 
The  question  is  sometimes  asked,  "  What  is  the  force  of  a 


54  EXAMPLES.  [28.1 

blow  of  a  hammer  ? "  If  by  the  force  is  meant  the  pressure 
hi  pounds  between  the  face  of  the  hammer  and  the  object 
struck,  it  cannot  be  determined  unless  the  law  of  resistance 
to  compression  between  the  bodies  is  known  during  the  con- 
tact of  tlie  bodies.  But  this  law  is  generally  unknown.  The 
pressure  begins  with  nothing  at  the  instant  of  contact  and 
increases  very  rapidly  up  to  the  instant  of  greatest  compression, 
after  which  the  pressure  diminishes.  The  pressure  involves 
the  elasticity  of  both  bodies ;  the  rapidity  with  which  the  force 
is  transmitted  from  one  particle  to  another ;  the  amount  of  the 
distortion ;  the  pliability  of  the  bodies ;  the  duration  of  the 
impact ;  and  some  of  these  depend  upon  the  degree  of  fixed- 
ness of  the  body  struck  ;  and  several  other  minor  conditions ; 
and  hence  we  consider  it  impossible  to  tell  exactly  what  the 
force  is. 

Examples, 

1.  Two  bodies  whose  weights  are  W  and  W^  are  placed 
very  near  each  other,  and  an  explosive  is  discharged  between 
them  ;  required  the  relative  velocities  after  the  discharge. 

2.  A  man  stands  upon  a  rough  board  which  is  on  a  perfectly 
smooth  plane,  and  jumps  off  from  the  board ;  required  the 
relative  velocities  of  the  man  and  board. 

[Obs.  The  common  centre  of  gravity  of  tlie  man  and  board  will  remain  the 
same  after  they  separate  that  it  was  before.  After  separating  they  would 
move  on  forever  if  they  did  not  meet  with  any  obstacle  to  prevent  their 
motion.] 

3.  A  man  whose  weight  is  150  pounds  walks  from  one  end 
of  a  rough  board  to  the  other,  which  is  twelve  feet  long,  and 
free  to  slide  on  a  perfectly  smooth  plane  ;  if  the  board  weighs 
60  pounds,  required  the  distance  travelled  by  the  man  in  space. 

4.  In  example  3  of  article  24,  suppose  that  the  weight  10 
pounds  is  permitted  to  fall  freely  through  a  height  A,  when  it 
produces  an  impulse  on  the  body  (50  pounds)  through  the 
intermediate  inextensible  string ;  required  the  initial  velocity 
of  the  body. 


(28.]  EXAMPLES.  55 

Let  Vq  ==  V2^A  =  the  velocity  of  the  weight  just  before 
.  the  impulse  ;  and 
V  =  the  velocity  immediately  afterward,  which  will 
be  the    common   velocity   of    the    body   and 
weight ; 

The  subsequent  motion  may  be  found  by  equation  (21), 
observing  that  the  initial  velocity  is  v. 

The  tension  on  the  string  will  be  infinite  if  it  is  inextensible, 
but  practically  it  will  be  finite,  for  it  will  be  more  or  less 
elastic. 

[Some  writers  have  used  the  expression  impulsive  tension  of  the  string 
instead  of  momentum.] 

5.  If  a  shell  is  moving  in  a  straight  line,  in  vacuo,  with  a 
velocity  v,  and  bui*sts,  dividing  into  two  parts,  one  part  moving 
directly  in  advance  with  double  the  velocity  of  the  body  ;  what 
must  be  the  ratio  of  the  weights  of  the  two  parts  so  that  the 
other  part  will  be  at  rest  after  the  body  bursts  ? 

6.  Explain  how  a  person  sitting  in  a  chair  may  move  across 
a  room  by  a  series  of  j6rks  without  touching  the  floor.  (Can 
he  advance  if  the  floor  is  perfectly  smooth  ?) 

7.  A  person  is  placed  on  a  perfectly  smooth  plane,  show 
how  he  can  get  off  if  he  cannot  reach  the  edge  of  the  plane. 


The  same  impulse  applied  to  a  small  body  will  impart  a 
greater  amount  of  energy  than  if  applied  to  a  large  one. 
Thus,  in  the  discharge  of  a  gun,  the  impulse  imparted  to  the 
gun  equals  that  imparted  to  the  ball,  but  the  work,  or  destruc- 
tive effect,  of  the  gun  is^small  compared  with  that  of  the  ball. 
The  ti?ne  of  the  action  of  the  explosive  is  the  same  upon  both 
bodies,  but  the  space  moved  over  by  the  gun  will  be  small 
compared  with  that  of  the  ball  during  that  time. 

The  product  Mv^  being  the  same  for  both,  as  J/"  decreases  v 
increases,  but  the  work  varies  as  the  square  of  the  velocity. 


66  IMPACT.  I29.J 

DIRECT    CENTRAL    IMPACT. 

29.  If  two  bodies  impinge  upon  one  another,  so  that  the  line 
of  motion  before  impact  passes  through  the  centre  of  the  bodies, 
it  is  said  to  be  central ;  and  if  at  the  same  time  the  common 
tangent  at  the  point  of  contact  is  perpendicular  to  the  line  of 
motion,  it  is  said  to  be  direct  and  central.     If  their  common 
tangent  is  perpendicular  to  the  line  of  motion,  but  if  the  latter 
does  not  pass  through  the  centre  of  the  body  impinged  upon, 
it  is  called  eccentric  impact.     In  this  place,  we  consider  only 
the  simplest  case ;  that  in  which  the  inipact  is  direct  and  central. 
When  two  bodies  impinge  directly  against  one  another,  whether 
r.^oving  in  the  same  or  opposite  directions,  they  mutually  dis- 
place the  particles  in  the  vicinity 
jp'  p  of  the  point  of  contact,  producing 

compression  which  goes  on  increas- 


ing until  it  becomes  a  maximum, 
at  which  instant  they  have  a  coni- 
\  ^       Fig.  17.  ^  YCiOTx  vclocity.    A  Complete  analysis 

of  the  motion  during  contact  in- 
volves a  knowledge  of  the  motion  of  all  the  particles  of  tlie 
mass,  and  wxaild  make  an  exceedingly  complicated  problem, 
but  the  motion  at  the  instant  of  maximum  corapj-ession  may  be 
easily  found  if  w^e  assume  that  the  compression  is  instantly  dis- 
tributed throughout  the  mass. 

Let  J/i  and  M^  be  the  respective  masses  of  the  bodies ; 

-^1  and  v^  the  respective  velocities  before  impact,  both 

positive  and  v^^v^ ; 
Vi  and  v^  the  respective  velocities  at  the  instant  of 

maximum  compression,  7/3  >  V2^  and 
§1  and   Q2  the  momenta  gained  respectively  by  the 
bodies  during  compression. 
Then  from  (31) 

which  will  be  the  momentum  gained  by  M^  on  account  of  the 
action  of  M-^. 
Similarly 

Q^  =  M,  {v^  -  ^•l), 

which,  being  negative,  will  be  the  momentum  lost  by  M^  on 
account  of  the  action  of  M^, 


[30.]  IMPACT.  57 

But  at  the  instant  of  greatest  conripression 

and,  because  they  are  in  mutual  contact  during  the  same  time, 
their  time-effects  are  equal,  but  in  opposite  directions, 

.-.  <2,  =  -  Q.- 

Combining  tliese  four  equations,  we  find  by  elimination 

which  velocity  remains  constant  for  perfectly  non-elastic  bodies 
after  impact,  since  such  bodies  have  no  power  of  restitution 
and  will  move  on  with  a  common  velocity. 

DIRECT    CENTRAL    IMPACT    OF    ELASTIC    BODIES. 

30.  Elastic  bodies  are  such  as  regain  a  part  or  all  of  their 
distortion  when  the  distorting  force  is  removed.  If  they  regain 
their  original  form  they  are  cdMed  perfectly  elastic,  but  if  only 
a  pai't,  they  are  called  imperfectly  elastic.  After  the  impact 
has  produced  a  maximum  compression,  the  elastic  force  of  the 
bodies  causes  them  to  separate,  but  all  the  effect  which  the 
force  of  restitution  can  produce  upon  the  movement  of  the 
bodies,  evidently  takes  place  while  they  are  in  contact.  If  they 
are  perfectly  elastic  and  do  not  fully  regain  their  form  at  the 
instant  of  separation,  they  will  continue  to  regain  their  form 
after  separation,  but  the  latter  effect  we  do  not  consider  in  this 
place.  The  ratio  between  the  forces  of  compression  and  those 
of  restitution  has  often  been  called  the  modulus  of  elasticity  ^ 
but  as  some  ambiguity  results  from  this  definition,  we  will  call 
it  the  modulus  of  restitution.  At  every  point  of  the  restitution 
there  is  assumed  to  be  a  constant  ratio  between  the  force  due 
to  compression  and  that  to  restitution.  But  it  is  unnecessary 
for  present  purposes  to  trace  these  effects,  for  by  equation  (31) 
we  may  determine  the  result  when  the  bodies  finally  separate 
from  each  other. 


58  IMPACT.  [31.] 

Let  «,  =  the  ratio  of  the  force  of  compression  to  that  of 
restitution  of  one  body,  which  is  called  the 
inodulus  of  restitution, 
€2  =  the  corresponding  value  for  the  other ; 
"Fj  =  the  velocity  of  M^^  at  the  instant  when  they  separ- 
ate from  each  other ;  and 
Y^  =  the  corresponding  velocity  for  Jfj. 

Then  from  equation  (31) 

e,  Q,  =  M,{Y,-v^y,  (34) 

e^Q^^M^{J^-v^\  (35) 

As  before  ^1  =  —  ^2  ^"^^  we  will  also  assume  that  ^^  =  <?2  =  A 
These  combined  with  (32)  and  (33)  give 

31.  Discussion  of  Equations  (36)  and  (37). 
1^.  If  the  bodies  are  perfectly  non-elastic,  <?  =  0. 

••   '^^-    M^  +  M^     ~  ^»'  /^^) 

which  is  the  same  as  (33). 

2°.  If  the  restitution  is  perfect  ^  =  1. 

F,=  ..,-f-;^^;^(^,-^,).  (40) 


[31.]  mPAOT.  59 

From  (38)  we  have 

and,  Y,-v,=  ^-^^^    K-tr,). 

Similarly  from  (39)  and  (40) 

■^'-^'=  =  1^+1^  ("'-*')' 

hence,  the  velocity  lost  by  one  body  and  gained  by  the  othei'  ia 
twice  as  much  when  the  bodies  are  perfectly  elastic  as  when 
they  are  perfectly  non-elastic. 

3°.  If  J/i  =  J/j,  then  for  perfect  restitution  we  have 

that  is,  they  will  interchange  velocities. 

4°.  If  Ml  impinges  against  a  fixed  body,  we  have  M^  =  oo , 
and  Vi  =  0. 

.-.   7i  =  -  ^  Vi . 

This  furnishes  a  convenient  mode  of  determining  e.  For  if 
a  body  falls  from  a  height  h  upon  a  fixed  horizontal  plane,  it 
will  rebound  to  a  height  h^  ; 

.'.  Ai  =  ^h,  or  e  =  \J  -^- 
Also  if  tf  =  1 

or  the  velocity  after  impact  will  be  the  same  as  before,  but  in 
an  opposite  direction. 


60 


EXAMPLES. 


[32.] 


Also  if  6  =  0,  Fj  =  0  ;  or  the  velocity  will  be  destroyed. 
6®.  If  ^2  =  0  we  have 


>  m 


Examples. 

(1.)  Amass  Mi  with  a  velocity  of  10,  impinges  on  M-2  moving  in  an  oppoeite 
direction,  moving  with  a  velocity  4  and  has  its  velocity  reduced  to  5  ;  required 
the  relative  magnitudes  of  Mi  and  Jfa. 

(2.)  Two  inelastic  bodies,  weighing  8  and  5  pounds  respectively,  move  in 
the  same  direction  with  velocities  7  and  3 ;  required  the  common  velocity 
after  impact,  and  the  velocity  lost  and  gained  by  each. 

(3.)  If  Ml  weighs  12  pounds  and  moves  with  a  velocity  of  15,  and  is  im- 
pinged upon  by  a  body  M^  weighing  16  pounds,  producing  a  common  velocity 
of  30,  required  the  velocity  of  Mi  before  impact  if  it  moves  in  the  same  or 
opposite  direction. 

(4.)  If  ^Mi  —  QMi ,  6oi  =  —  6«a ,  «2  =  7,  and  e  =  | ;  required  the  velocity 
of  each  after  impact. 

(5.)  If  Ml  =  2Mi ,  Vi  —  |€i,  and  ^2=0;  required  e. 

(6.)  If  ?)i  is  26,  Ml  is  moving  in  an  opposite  direction  with  a  velocity  of  16  ; 
Ml  =  2M-1 ,  «  =  I ;  required  the  distance  between  them  5^  seconds  after 
impact. 

(7.)  Two  bodies  are  perfectly  elastic  and  move  in  opposite  directions ;  the 
weight  of  Ml  is  twice  Mi^  but  n-i  =  2©i ;  required  the  velocities  after  impact. 

(8.)  There  is  a  row  of  perfectly  elastic  bodies  in  geometrical  progression 
whose  common  ratio  is  3,  the  first  impinges  on  the  second,  the  second  on  the 
third  and  so  on  ;  the  last  moves  off  with  -^  the  velocity  of  the  first.  What  ii 
the  number  of  bodies  ? 

Am.  7. 


LOSS   OF   VIS    VrVA  IN  THE    IMPACT   OF   BODIES. 

32,  Before  impact  the  vis  viva  of  both  bodies  was 
and  after  impact 


[33.]  RELATIONS  OF  FORCE,  MOMENTUM,  AND  WORK.  61 

which  by  means  of  (36)  and  (37)  becomes 

M^  Y?  +  M,  Yi  =  M,v}  +  M,vi  -  ^^|-^^(,;,_^,)l  (42) 

For  perfectly  elastic  bodies  6  =  1  and  the  last  term  disap- 
pears ;  hence  in  the  imjpact  of  jperfectly  elastic  bodies  no  vis 
viva  is  lost. 

If  the  bodies  are  imperfectly  elastic  e  is  less  than  1,  and  smce 
{qj^  __  q)^^  is  always  positive,  it  follows  that  in  the  imjpact  of 
imperfectly  elastic  hodies  vis  viva  is  always  lost^  and  tJie 
greatest  loss  is  suffered  when  the  hodies  are  perfectly 
non-elastic. 

If  ^  =  0,  (42)  becomes 

M,W  -  V?)  +  M,{vi  -  Vr)  =-^^^  K  -  ^.f  ;    (43) 

in  which  each  member  is  the  total  loss  by  both  bodies.  It  is 
also  the  loss  up  to  the  instant  of  greatest  compression  when  the 
bodies  are  elastic. 

If  il^  is  very  large  compared  with  M^  we  have  from  (38) 
Yi  =  v^  nearly,  =  Fg , 
and  (43)  becomes 

the  second  member  of  which  is  frequently  used  in  hydraulics 
for  finding  the  vis  viva  lost  by  a  sudden  change  of  velocity. 

These  investigations  show  the  great  utility  of  springs  in 
vehicles  and  machines  which  are  subjected  to  impact. 


RELATIONS   OF   FOEOE,   MOMENTUM,   WORK,   AND   VIS   VIVA. 

38,  We  may  now  determine  the  exact  office  in  the  same 
problem  of  the  quantities  \—force^  momentum,  work,  and  via 
viva.  Suppose  that  a  force,  whether  variable  or  constant, 
impels  a  body,  it  will  in  a  time  t  generate  in  the  mass  M  a 
certain  velocity  v.  Th.\%  force  may  at  any  instant  of  its  action 
be  measured  by  a  certain  number  of  pounds  or  its  equivalent. 


62  IMPULSE.     STATICS.  [34,  35.  j 

Suppose  that  this  mass  impinges  upon  another  body,  which  may 
be  at  rest  or  in  motion.  In  order  to  determine  the  effect  upon 
their  velocities  we  use  the  principle  of  inomentum^  as  has  been 
shown.  But  the  bodies  are  compressed  during  impact  and 
hence  work  is  done.  The  amount  of  work  which  they  are 
capable  of  doing  is  equal  to  the  sum  of  tlieir  vis  viva  /  and  if 
they  are  brought  to  rest  all  this  work  is  expended  in  compress- 
ing them.  If  the  velocity  of  a  body  after  impact  is  less  than 
that  before,  it  has  done  an  amount  of  work  represented  by 
\M{v^  —  F^),  and  similarly  if  the  other  body  has  its  velocity 
increased  kinetic  energy  is  imparted  to  it.  The  distortions  of 
lodies  represent  a  certain  amount  of  work  expended.  And  this 
explains  why  in  the  impact  of  imperfectly  elastic  bodies  vis 
viva  is  always  lost,  for  a  portion  of  the  distortion  remains.  But 
no  force  is  lost.  One  of  the  grandest  generalizations  of  physical 
science  is,  that  no  energy  in  nature  is  lost.  In  the  case  of  im- 
pact, compression  develops  heat,  and  this  passes  into  the  air  or 
surrounding  objects,  and  the  amount  of  energy  which  is  stored 
in  the  heat,  electricity  or  other  element  or  elements,  which  is 
developed  by  the  compression,  exactly  equals  that  lost  to  the 
masses.  We  thus  see  that  in  the  case  of  moving  bodies,  force 
impels^  momentu'm  determines  velocity  after  impact,  aoid  work 
or  vis  viva  represents  the  resistance  which  the  jparticles  offer  to 
heing  displaced, 

34.  Statics  is  that  case  in  which  the  force  or  forces  which 
would  produce  motion  are  instantly  arrested,  resulting  in  pres- 
sure only.  -The  expression  for  the  elementary  work  which  a 
force  can  do  is  Fds,  but  if  the  space  vanishes,  we  have,  Fds  =  0. 
This,  as  we  shall  see  hereafter,  is  a  special  case  of  "  virtual 
velocities." 

The  forces  which  act  upon  a  body  may  be  in  equilibrium  and 
yet  motion  exist,  but  in  such  cases  the  velocity  is  uniform. 

35.  The  term  power  is  often  used  in  the  same  sense  sls  force, 
but  generally  it  refers  to  an  acting  agent.  The  term  mechan- 
ical power  is  not  only  recognized  in  this  science,  but  has  a 
specific  meaning,  and  for  the  purpose  of  avoiding  ambiguity, 
it  is  better  to  use  the  term  effort  in  reference  to  mechanical 
agents.  Thus,  instead  of  saying  the  pow&i^  and  weight,  as  is 
often  done,  say  the  effort  and  resistance.  • 


[36-38.]  INERTIA.  '  63 

36.  Inertia  implies  passiveness  or  want  of  power.  It  means 
that  matter  has  no  power  within  itself  to  put  itseK  in  motion, 
or  when  in  motion  to  change  its  rate  of  motion.  Unless  an 
external  force  be  applied  to  it,  it  would,  if  at  rest,  remain  for- 
ever in  that  condition ;  or  if  in  motion,  continue  forever  in 
motion.  Gravity,  which  is  a  force  apparently  inherent  in  mat- 
ter, can  produce  motion  only  by  its  action  upon  other  matter. 

Inertia  is  not  a  force^  but  because  of  the  propei-ty  above 
explained,  those  impressed  forces  which  produce  motion  are 
measured  by  the  product  of  the  mass  into  the  acceleration  as 
explained  in  preceding  articles ;  and  many  writers  call  this 
MEASURE  the  force  of  inertia. 

37.  Newton's  Three  Laws  of  Motion. 

Sir  Isaac  Newton  expressed  the  fundamental  principles  of 
motion  in  the  form  of  three  laws  or  mechanical  axioms;  as 
follows : — 

1st.  Every  body  continues  in  its  state  of  rest  or  of  uniform 
motion  in  a  straight  line  unless  acted  upon  by  some  externa] 
force. 

2d.  Change  of  motion  is  proportional  to  the  force  impressed, 
and  is  in  the  direction  of  the  line  in  which  the  force  acts. 

3d.  To  every  action  there  is  opposed  an  equal  reaction. 

[As  simple  as  these  laws  appear  to  the  student  of  the  present  day,  the 
science  of  Mechanics  made  no  essential  progress  until  they  were  recognized. 
See  Whewell's  Inductive  Sciences^  3d  ed.,  vol.  1,  p.  311.] 

38.  In  all  the  problems  thus  far  considered,  it  has  been 
assumed  that  the  ac- 
tion-line of  the  force  or  ,    "'  "V'-v 

forces  passed  tlirough  ..-  'j^^i^j^ P  -  (^^ )  +  v"^ 

the  centre  of  the  mass, 
producing  a  motion  of 
translation  only.  But 
if  the  action-line  does 
not  pass  through  the 
centre,  it  will  produce 
both  translation    and    {^-^  ) — ' — ^lilSiN 


64  DIHECT  ECCENTRIC  IMPACT.  [38.] 

In  eccentric  hnjpact  both  translation  and  rotation  is  pro- 
duced. The  centre  of  the  body  will  move  in  a  straight  line, 
hut  every  other  jpoiiit  will  describe  arcs  of  circles  in  reference 
to  the  centre  of  the  body^  which  in  space  will  be  curves  more  or 
less  elongated.  The  velocity  of  traixslation  will  be  directly 
proportional  to  the  intensity  of  the  impulse  imparted  to  the 
body^  bid  the  angular  velocity  will  depend  upon  the  intensity 
of  the  impulse  and  the  distance  of  the  point  of  impact  from 
the  centre  of  the  body. 

In  Figure  18,  let  Q  =  Mv  be  the  impulse  imparted  to  the 
body ;  in  which  M  is  the  mass  of  the  body  and  'y  the  velocity 
of  the  centre.  Let  this  impulse  be  imparted  at  a.  At  b,  a 
distance  from  the  centre  =  cb  =  ac,  let  two  equal  and  opposite 
impulses  be  imparted,  each  equal  to  ^Q.  The  impulse  Q, 
equals  -kQ  +  iQ-  The  four  impulses  evidently  produce  the 
same  effect  upon  the  body  as  the  single  impulse  Q.  If  now 
one  of  the  impulses,  ^Q,  above  the  centre  is  combined  with 
the  equal  and  parallel  one  acting  in  the  same  direction  below 
the  centre,  their  effect  will  be  equivalent  to  a  single  one,  equal 
to  Q  applied  at  the  centre  c.  This  produces  translation  only. 
The  other  -J  Q  above  the  centre  combined  with  the  equal  and 
opposite  ^Q  below  the  centre,  produces  rotation  only ;  and  it 
is  evident  that  the  greater  the  distance  a,  the  point  of  impact, 
is  from  the  centre,  the  greater  will  be  the  amount  of  rotation. 

An  impact  (or  blow)  at  a  to  produce  a  velocity  v  at  the  centre 
of  the  body,  must  act  through  a  greater  space  during  contact, 
or  the  impacting  body  must  move  with  a  greater  velocity,  than 
if  the  impact  be  in  a  line  passing  through  the  centre  c. 

(The  entire  energy  stored  in  the  body  will  be  ^Mv^  +  iim^^,  in  which  Im  is 
the  moment  of  inertia  of  the  rotating  mass  in  reference  to  an  axis  through  the 
centre,  and  w  is  the  angular  velocity  in  reference  to  the  same  axis  ;  and  the 
other  notation  is  the  same  as  in  the  preceding  Article.  See  Article  127.  This 
expression  for  the  energy,  in  case  the  bodies  are  perfectly  elastic,  will  equal 
the  energy  lost  by  the  impacting  body.) 


CHAPTEK  11. 
COMPOSITION  AND  RESOLUTION  OF  FORCES. 

CONCURRENT   FORCES. 

39,  If  two  or  more  forces  act  upon  a  material  particle,  they 
are  said  to  be  concurrent.  They  may  all  act  towards  the 
particle,  or  from  it,  or  some  towards  and  others  from. 

40,  If  several  forces  act  along  a  material  line,  they  are 
called  conspiring  forces,  and  their  effect  will  be  the  same  as  if 
all  were  applied  at  the  same  point. 

41,  The  Resultant  of  two  or  more  concurrent  forces  is  that 
force  which  if  substituted  for  the  system  will  produce  the 
same  effect  upon  a  particle  as  the  system. 

Therefore,  if  a  force  equal  in  magnitude  to  the  resultant  and 
acting  along  the  same  action-line,  but  in  the  opposite  direction, 
be  applied  to  the  same  particle,  the  system  will  be  in  equilib 
rium. 

If  the  resultant  is  negative,  the  equilibrating  force  will  be 
positive,  and  vice  versa. 

Hence,  if  several  concurrent  forces  are  in  equilibrium,  any 
one  may  be  considered  as  equal  and  opposite  to  the  resultant 
of  all  the  others. 

42,  The  resultant  of  several  conspiring  forces,  equals  the 
algebraic  sum  of  the  forces.  That  is,  if  i^ ,  i^  ,  J^ ,  etc.,  are  the 
forces  acting  along  the  same  action-line,  some  of  which  may  be 
positive  and  the  others  negative,  and  R  is  the  resultant ;  then 

^  =  i^+i^  +  i^  +  etc.  =  XF,  (45) 

43,  If  two  concurring  forces  he  represented  in  magnitude 
and  direction  hy  th^  adjacent  sides  of  a  parallelogram,,  the 
resultant  will  he  represented  in  magnitude  and  direction  hy 
the  diagonal  of  the  parallelogram.  This  is  called  the  paral- 
lelogram of  forces. 

If  each  force  act  upon  a  particle  for  an  element  of  time 
it  will  generate  a  certain  velocity.     See  equation  (44).     Lei 


m 


PARALLELOGRAM  OP  FORCES. 


[48.1 


the  velocity  which  i^ would  produce  be  represented  by  AB  ; 
and   that  of  P  by  AD  =  jBO.      These  represent   the  spaces 

over  which  the  forces 
respectively  would  move 
the  particle  in  a  unit  of 
time  if  each  acted  sepa- 
rately. If  we  conceive 
that  the  force  jp^  moves 
it  from  AtoB  and  that 
the  motion  is  there  ar- 
Fio.  u.  rested,   and    that   P    is 

then  applied  at  B,  but  acthig  parallel  to  AD,  then  will  the 
particle,  at  the  end  of  two  seconds,  be  at  O.  If,  next,  we  con- 
ceive that  each  force  acts  alternately  during  one-half  of  a 
second  beginning  again  at  A,  the  particle  will  be  found  at  a  in 
one-half  of  a  second  ;  at  J  at  the  end  of  one  second ;  at  c  at  the 
end  of  one  and  one-half  seconds  ;  and  finally  at  O  at  the  end 
of  two  seconds.  If  the  times  be  again  subdivided  the  path  will 
be  Ad,  de,  ef.fb,  hg,  gh,  hi,  and  iC,  and  it  will  arrive  at  C  in 
the  same  time  as  before. 

As  the  divisions  of  the  time  increase,  the  number  of  sides  of 
the  polygon  increase,  each  side  becoming  shorter;  and  the 
polygonal  path  approaches  the  straight  line  as  a  limit.  There- 
fore at  the  limit,  when  the  force  P  and  i^act  simultaneously, 
the  particle  will  move  along  the  diagonal,  AO,  of  the  parallelo- 
gram. But  when  they  act  simultaneously,  they  will  produce 
their  effect  in  the  same  time  as  each  when  acting  separately ; 
and  hence,  the  particle  will  arrive  at  C  at  the  end  of  one  second. 
Therefore,  a  single  force  R,  which  is  represented  by  A  C,  will 
produce  the  same  effect  as  P  and  F,  and  will  be  the  resultant. 
If  now  a  force  equal  and  opposite  to  E  act  at  the  same  point 
as  the  forces  i^and  P,  the  motion  will  be  arrested  and  pressure 
only  will  be  the  result.  See  article  34.  Hence,  the  parallelo- 
gram of  velocities  and  of  pressures  becomes  established.* 


*  This  is  one  of  the  most  important  propositions  in  Mechanics,  and  has 
been  proved  in  a  variety  of  ways.  One  work  gives  forty-five  different  proofs, 
A  demonstration  given  by  M.  Poisson  is  one  of  the  most  noted  of  the  analytical 
proofs.     Many  persons  object  to  admitting  the  idea  of  motion  in  proving  the 


[44.J 


TRIANGLE  OP  FORCES. 


67 


If  Q  be  the  angle  between  the  sides  of  the  parallelogram 
which  represent  tlie  forces  P  and  F^ 
and  M  be  the  diagonal,  or  resultant, 
we  have  from  trigonometry 


^  =  i^  +  P^  +  ^PF  cos  e,     (46) 

If  0  exceeds  90  degrees,  it  must  be 
observed  in  the  solution  of  problems 
that  cos  B  will  be  negative. 

If  ^  =  90  degrees,  we  have 


>R 


Fio.  90. 


Fia.  21 


Also,  if  ^  =  90°,  and  a  be  the  angle  between  J?  and  P\ 
and  |S  between  R  and  F\  then 


(47) 


P  =  P  coBa;  ) 

F  =  P  coa  §  =  P  sina.  J 

Squaring  and  adding,  we  have 

P'  +  F'  =  P', 
as  before. 

The  forces  P  and  i^are  called  component  Jbrces, 

44.  Triangle  of  Forces.  If  three  concurrent  forces  are  in 
equilibrium,  they  may  he  represented  in  magnitude  and  direc- 
tion hy  the  sides  of  a  triangle  taken  in  their  order  ;  and  if  the 
direction  of  action  of  one  he  reversed,  it  will  he  the  resultant 
of  the  other  two. 

Thus,  in  Fig.  19,  if  AB  and  BC  represent  two  forces  in 
magnitude  and  direction,  A  C  will  repi-esent  the  resultant. 

parallelogram  of  pressures ;  but  we  have  seen  that  a  pressure  when  acting 
upon  a  free  body  will  produce  a  certain  amount  of  motion,  and  that  this  motion 
is  a  measure  of  the  pressure,  and  hence  its  use  in  the  proof  appears  to  be 
admissible.  But  the  strongest  proof  of  the  correctness  of  the  proposition  ia 
tho  fact  that  in  all  the  problems  to  which  it  has  been  applied,  the  resultt 
agree  with  those  of  experience  and  observation. 


68 


POLYGON  OF  FORCES. 


[45-46.1 


Since  the  sines  of  the  angles  of  a  triangle  are  proportional 
to  the  sides  opposite,  we  have 


R 


A  A 

sinP,^    sini^^ 


sini^ 


(48) 


POLYGON   OF   FORCES. 

45.  If  several  ioiv 
current  forces  are  re- 
presented in  magni- 
tude and  direction  hy 
the  sides  of  a  closed 
polygon  taken  in  their 
order,  they  will  he  in 
equilibrium. 

This  may  be  proved  bj  finding  the  resultant  of  two  forces 
by  means  of  the  triangle  of  forces  ;  then  the  resultant  of  that 
resultant  and  another  force,  and  so  on. 


Fio  21a. 


PARALLELOPIPED  OF   FORCES. 

46.  If  three  concurrent  forces  not  in  the  same  jplane  are 
represented  in  magnitude  and  direction 
hy  the  adjacent  edges  of  a  jparallelopijpe- 
don,  the  resultant  will  he  represented  in 
magnitude  and  directio/i  hy  the  diago- 
nal ;  and  conversely  if  the  diagonal  of  a 
jparallelo^ipedon  represents  .  a  force,  it 
may  he  considered  as  the  resultant  of 
three  forces  represented  hy  the  adjacent 
edges  of  the  p>arallelojpi/pedon. 

In  Fig.  22,  if  AD  represents  the  force  Fi  in  magnitude  and 
direction,  and  similarly  DB  represents  i^,  and  BC,  Fz\  then 
according  to  the  triangle  of  forces  AB  will  represent  the 
resultant  of  F^  and  F^ ;  and  A  O  the  resultant  of  AB  and  F^ 
and  hence  it  represents  the  resultant  of  i^,  i^,  and  F^, 


Fio. 


[4«.]  EXAMPLES.  69 

If  Fi,  F2,  and  i^  are  at  right  angles  with  each  other,  we 
have 

and  if  «  is  the  angle  ByFi ,  §  of  ^^Fl ,  and  7  of  -S,i^ ;  then 

i^  =  ^  cos  a ;    \ 

F2  =  Iicos^;    i  (49) 

i^  =  ^  cos  y.     ) 

Squaring  these  and  adding,  we  have 

l^  =  F?  +  Fi-\-F^\2^  before. 

Examples. 

1.  When  F=  7^  and  ^  =  60°,  find  E ;  (See  Eq.  (46) ). 

Ans.  R  =  FVS, 
2.UF=F,  and  0  =  120°,  find  R. 

3.  If  F=  F,a.nd0  =  135^,  find  B. 

Ans,  R  =  F\/¥^y/^^ 

4.  n  i^=  27^  =  3^,  find  ^. 

6.  If  \F^  Ji^i  =  7^,  find  the  angle  F,F^, 

Ans.  90° 

6.  If  7^=  r,  7^  =  9,  and  0  =  25°,  find  ^  and  angle  F^R, 

7.  A  cord  is  tied  around  a  pin  at  a  fixed  point,  and  its  two 
ends  are  drawn  in  different  directions  by  forces  7^ and  P.  Find 
0  when  the  pressure  upon  the  pin  is  ^  =  -J  (P  4-  7^'). 

8.  When  the  concurring  forces  are  in  equilibrium,  prove  that 

AAA 
PiFiRiism  F,R  :  sin  F,R  :  sin  F,F 

9.  If  two  equal  rafters  support  a  weight  W  at  their  upper 
ends,  required  the  compression  on  each.     Let  the  length  of 


70 


OONCURBENT  FORCES. 


[47.1 


each  rafter  be  a  and  the  horizontal  distance  between  their  lower 
ends  be  5. 

Ans.       .— 


V4a'-l^ 


W, 


10.  If  a  block  whose  weight  is  200  pounds  is  so  situated 
that  it  receives  a  pressure  from  the  wind  of  25  pounds  in  a 
due  easterly  direction,  and  a  pressure  from  water  of  100 
pounds  in  a  due  southerly  direction ;  required  the  resultant 
pressure  and  the  angle  which  the  resultant  makes  with  the 
vertical. 


RESOLUTION   OF   CONCURRENT   FORCES. 

47*  I^et  there  be  many  concurrent 
forces  acting  upon  a  single  particle, 
and  the  whole  system  be  referred  to 
,        rectangular  co-ordinates. 

Let  i^I,i^,  i^,  etc.,  be  the  forces 
acting  upon  a  particle  at  A ; 
X,  y,  z  the  co-ordinates  of  A  ; 
«!,  agj  etc.,  the  angles  which  the 
Fig.  23.  dircction-liues  of  the  respec- 

tive forces  make  with  the  axis  of  x ; 
ft  5  ft )  etc.,  the  angles  which  they  make  with  y  ; 
yi ,  72  >  etc.,  the  angles  which  they  make  with  z ;  and 
Z",  T',  and  Z,  the  algebraic  sum  of  the  components  of  the 
forces  when  resolved  parallel  to  the  axes  »,  y,  and  2, 
respectively. 

Then,  according  to  equations  (45)  and  (49),  we  have  for 
equilibrium  ; 

X—  i^costti+i^cosag-fi^cos  «3  +  etc.=2'  Fao^a  =  0  ;  j 
]r=i^  cos  ^i+i^cosft-hi<8«osft+etc.=:2'i^cos^=  0;  t  (50) 
Z  =  i^  cos  yi+Fz  cos  j'2+i^  cos  y^  +  etc.=-2'i^cos  y  =  0  ;  ) ' 

If  they  are  not  in  equilibrium,  let  H  be  the  resultant,  and 
by  introducing  a  force  equal  and  opposite  to  the  resultant,  the 
system  will  be  in  equilibrium. 


[48.] 


CONSTRAINED  EQUILIBRIUM, 


n 


Let  a,  h  and  c  be  the  angles  which  the  resultant  makes  with 
the  axes  a?,  y  and  2  respectively ;  then 
X  =  R  cos  a ; 
Z  =  i?cos&;  (51) 

Z  =  i?  cos  (7  . 

Squaring  and  adding,  we  have 

X^  +  r^  H-  Z^  =  i?»  (52) 

If  ^  =  0  equations  (51)  reduce  to  (50). 

When  the  forces  ar6  in  equilibrium  any  one  of  the  F -forces 
may  be  considered  as  a  resultant  (reversed)  of  all  the  others. 
Equations  (50)  are  therefore  general  for  concurring  forces. 

The  values  of  the  angles  a,  ^,  7,  etc.,  may  be  determined  by 
drawing  a  Xwi^frorrh  the  origin  parallel  to  and  in  the  direction 
of  the  action  of  the  force,  and  measuring  the  angles  from  the 
axes  to  the  line  as  in  Analytical  Geometry.  The  forces  may 
always  be  considered  as  positive,  and  hence  the  signs  of  the  terms 
in  (50)  will  be  the  same  as  those  of  the  trigonometrical  func- 
tions. In  Fig.  23  the  line  Oa  is  parallel  to  i^ ,  and  the  corre- 
sponding angles  which  it  makes  with  the  axes  are  indicated. 

If  all  the  forces  are  in  the  plane  x  y  then  71 ,  yg  >  etc.  =  90°, 
and  (50)  becomes 

X  =  Z  Fco^  «  =  0  ;    )  ,Ko\ 


CONSTRAINED   EQUILIBRIUM. 

48.  A  body  is  constrained  when  it  is  prevented  from  moving 
freely  under  the  action  of  applied  forces. 

If  a  particle  is  constrained  to  remain 
at  rest  on  a  surface  under  the  action  of 
any  number  of  concurring  forces,  the 
resultant  of  all  the  applied  forces  must 
be  in  the  direction  of  the  normal  to  the 
surface  at  that  point. 

For,  if  the  resultant  were  inclined  to 
the  normal,  it  could  be  resolved  into 
two  components,  one  of  which  would  be  ^w-  ^ 


72  CONSTRAINED  EQUILIBEIUM.  [48.] 

tangential,  and  would  produce  motion ;  and  the  other  normal, 
wliich  would  be  resisted  by  the  surface. 

Let  iV=  the  normal  reaction  of  the  surface,  which  will  be 
equal  and  opposite  to  the  resultant  of  all  the 
impressed  forces ; 
6^  =  the  angle  {JSF^x) ; 
6y  =  the  angle  {J^,y) ; 
6^  =  the  angle  (ir,s)  ; 
Z  =  <l){x,  y,  b)  =  0,  be  the  functional  equation  of  the 

surface;  and 
Fl,  JPl,  JF^,  etc.,  be  the  impressed  forces. 
Then  from  (51)  and  (52),  we  have 


X  =  JSr cos  0:^;  ] 

I^^iV'cos^y;  I 

Z=i\^cos^,;  .'' 


From  Calculus  we  have 
cos^~   = 


(54) 


(55) 


w_/ 

and  similarly  for  cos  dy  and  cos  6^ . 
These  values  in  (54)  readily  give 

X    _  ^Y^  _  _Z^ 
\dx)       \dyl       [(12/ 
After  substituting  the  values  of  cos  Og, ,  cos  6y ,  and  cos  0^  in 


(56), 


[48.1 


CONSTRAINED  EQUILIBRIUM. 


73 


(54),  multiply  the  first  equation  by  dx^  the  second  by  dy^  the 
third  by  <is,  add  the  results,  and  reduce  by  the  equation 

(g^.  (I)  .,.(«).,  =  »; 

which  is  the  total  differential  of  the  equation  Z  =  0  ;   and  we 
have 

^dx  ^-  Ydy  +  Zdz  =  0.  (57) 

Equations  (56)  give  two  independent  simultaneous  equations 
which,  combined  with  the  equation  of  the  surface,  will  deter- 
mine the  point  of  equilibrium  if  there  be  one.  Equation  (57) 
is  one  of  condition  which  will  be  satisfied  if  there  be  equilibrium. 

To  deduce  (55)  let  /  («',  y')  =  0,  and  /'  (a^, 
e')  =  0,  be  the  equations  of  the  normal  to  the 
surface  at  the  point  where  the  forces  are  applied. 
In  Fig.  25  let  Oa  be  drawn  through  the  origin  of 
co-ordinates  parallel  to  the  required  normal, 
then  wiU  dx\  dy'  and  dz'  be  directly  propor- 
tional to  the  co-ordinates  of  a  ; 


cos  aOx  =  cos  ex  = 


Oa 


dx 


Fig.  25. 


y/da^+dy'-'+dz" 


But  the  normal  is  perpendicular  to  the  tangent  plane,  and  hence  the  pro- 
jections of  the  normal  are  perpendicular  to  the  traces  of  the  tangent  plane. 
The  Equation"  of  Condition  of  Perpendicularity  is  of  the  form  14-  aa'  —0 

(See  Analytical  Geometry)  ;   in  which  a'  =  ^ ,  and  a  =  ^  ;    the  latter  of 

docf  dx 

which  is  deduced  from  the  equation  of  the  surface  ; 


*•  ^  ■•"  d^c'l^      ~  ^'  ^^  similariy 


1-1- 


dz'  dz 
dx'  dx 


74 

hence 


CONSTRAINED  EQUILIBRIUM. 


dx 


dx 


dz' 

TdH  ^  ^^  ^  = 


dz  '~  TrfTT  * 
[dx) 


[49.1 


the  last  terms  of  which  contain  the  partial  differential  co-efficients  deduced 
from  the  equation  of  the  surface.  These,  substituted  in  the  value  of  cos  Ox 
above,  and  reduced,  give  equation  (55). 


CONSTRAINED   EQUILIBRIUM  IN  A   PLANE. 

49-  If  all  the  forces  are  in  the  plane  of  a  curve,  let  the 
plane  yx  coincide  with  that  plane ;  then  Z  =  0  and  (56) 
becomes 


\dx  I      \dy  / 


(68) 


or,  Xdx  =  —  Ydy ; 

and,  Xdx  +  Ydy  =  0;  (59) 

in  which  the  first  of  (58)  may  be  used  when  the  equation  of 
the  curve  is  given  as  an  implicit  function ;  and  the  second  of 
(68),  or  (59),  when  the  equation  is  an  explicit  function. 

When  the  particle  is  not  constrained  it  has  three  degrees  of 
freedom  (equations  (50) )  ;  when  confined  to  a  surface,  two 
degrees  (equations  (56)) ;  and  when  confined  to  a  plane  curve, 
only  one  degree  (equation  (58)  ). 


Examples. 

1.  A  body  is  suspended  vertically  by  a*  cord  which  passes 
over  a  pulley  and  is  attached  to  another  weight  which  rests 

upon  a  plane;    required  the  position 
of  equilibrium. 

In  Fig.  26,  let  the  pulley  be  at  the 

upper  end  of  the  plane  and  the  cord 

and    plane    perfectly  smooth.      The 

Pia,  28w  weight    JP  is  equivalent  to  a  force 


[40.1 


EXAMPLES. 


75 


which  acts  parallel  to  the  plane,  tending  to  move  the  weight 
W  up  it. 

Let  W  =  the  weight  on  the  plane,  which  acts  vertically 
downwards ; 
P  =  the  weight  suspended  by  the  cord  ; 
1  =  the  inclination  of  the  plane  to  the  horizontal ;  and 
Z  =  —  y  +  ax  +  b  =  i),  he  the  equation  of  the  plane. 
Then  X  =  P  cos  * ; 

Y=  -TT  +  Psini; 

sin  i 
a  =  tan  z  = ; ; 

cos* 


(S  =  -^'"^^(S  =  ^' 


[dy 

and  these  in  (58)  give 

P  =  Trsin*; 

which  only  establishes  a  relation  between  the  constants,  and 
thus  determines  the  relation  which  must  exist  in  order  that 
there  may  be  equilibrium ;  and  since  the  variable  co-ordinates 
do  not  appear,  there  will  be  equilibrium  at  all  points  along  the 
plane  when  P  =  TFsin  i. 

The  equation  of  the  line,  given  explicitly,  is 

y  z=  ax  +  h; 
.\  dy  =  a  dx] 

which  in  the  2°"^  of  (58),  or  in  (59),  gives,  J*  =  TF  sin  i  as  before. 

2.  Two  weights  F  and  W  are  fastened  to  the  ends  of  a  cord, 
which  passes  over  a  pulley  0 ;  the  weight  W  rests  upon  a 
vertical  plane  curve,  and  P  hangs  freely  ; 
required  the  position  of  equilibrium. 

The  applied  forces  at  W  are  the  weight 
W,  acting  vertically  downward ;  the  ten- 
sion P  on  the  string ;  and  the  normal 
reaction  of  the  curve. 

[Consider  the  weight  W  and  pulley  0  as  re- 
duced to  points.  J 


Take  the  origin  of  co-ordinates  at  the 


FiQ.  «T. 


76  EXAMPLES.  [49.] 

pulley,  y  vertical  and  positive  upwards,  and  x  positive  to  the 
right.  The  angle  between  +  x  and  P  is  dbo ;  between  +  y 
and  P,  dfe.  Let  AW  =  x,  OA  =  y,  0W=  r,  WOa  =  0 ; 
then 

sin  ^  =  ^ ,  cos  ^  =  -,  7^  =  {»2  +  2/2 . 

X  =  TF  cos  270°  +  P  cos  ahG  =  0  •\-  P  cos  (180°  +  ff) 

=z  -  p  COS  e, 

T=  IF  sin  270°  +  P  sin  (180°  +  6) 
=  -W-  P  sin  0  ; 

and  (59)  becomes 

-  Pcosedx-{W-h  Psme)dy  =  0; 
or 

.Wdy  =  F^-^^!^^  =  Fdr.  (a) 

3.  Let  the  given  curve  be  a  parabola,  in  which  the  origin 
and  pulley  are  at  the  focus,  the  axis  vertical,  and  the  vertex  of 
the  curve  above  the  origin. 

The  equation  of  the  curve  will  be 

o^  =  2p{-y  +  ip), 

in  which  2p  is  the  principal  parameter.     (See  Analyt  Geo7n.) 
Differentiating  gives 

xdx  =  —J}dy, 

which  substituted  in  (a)  gives 

or 

W—  +  P  —      ^  ~^  —  4-  P- 

which  simply  establishes  a  relation  between  constants;  and 
therefore,  if  they  are  in  equilibrium  at  any  point  they  will  be 
at  evei-y  point,  and  there  w^ill  be  no  equilibrium  unless'  the 
weights  are  numerically  equal. 

4.  Let  the  curve  be  a  circle  in  w^hich  the  origin  and  pulley 
are  at  a  distance  a  above  the  centre  of  the  circle. 


[49.]  EXAMPLES.  77 

Since  y  is  negative  downwards,  we  have  for  the  equation  of 

the  circle, 

{a  +  yf  4-  ar^  =  ^. 

Differentiating  gives 

xdx  =  —  {a  -\-  y)  dy, 

which  substituted  in  {a)  gives 

P 

6.  Let  the  curve  be  art  hyperbola  having  the  origin  and  pulley 
at  the  centre  of  the  hyperbola,  the  axis  of  the  curve  being 
vertical. 

The  equation  of  the  curve  will  be  b^y^  —  aV  =  aW,  and  if 


i;^  -, 


€  be  the  eccentricity,  we  find  v-'^^^^ 

6.  Required  the  curve  such  that  the  weight  TFmay  be  in 
equilibrium  with  the  weight  P  at  all  points  of  the  curve. 

This  relation  requires  that  the  relation  between  y  and  x  in 
equation  (a)  shajl  be  true  for  all  assumed  values  of  P  and  W. 
We  have 

^^ ,         ^  xdx  +  ydy 

Integrating  gives 

-%+  c^py/x'^y". 

Squaring 

Fy  -  267%  -t-  (7^  ^  P^^  +  P'y', 
or 

p2a52  +  (P2  _  TT')  y»  +  ^CWy-  0^  =  0; 

which  is  an  equation  of  the  second  degree,  and  hence  repre- 
sents a  conic. 

If  P  =  IF,  it  is  a  parabola. 

If  P  >  W,  it  is  an  ellipse. 

If  P  <  TT,  it  is  a  hyperbola. 
The  origin  is  at  the  focus. 

7.  A  particle  is  placed  on  the  concave  surface  of  a  smooth 
sphere  and  acted  upon  by  gravity,  and  also  by  a  repulsive 


78  EXAMPLES.  [49.] 

force,  which  varies  inversely  as  the  square  of  the  distance  from 
the  lowest  point  of  the  sphere ;  find  the  position  of  equilibrium 
of  the  particle. 

Take  the  lowest  point  of  the  sphere  for  the  origin  of  coordi- 
nates, y  positive  upwards,  and  the  equation  of  the  surface 
will  be 

L  =  x^-\-y^  +  ^-  "^Ry  =  0. 

Let  T  be  the  distance  of  the  particle  from  the  lowest  point ; 
then 

7»  =  ;^  ^if  ^r^-  'i'Ry^  (b) 

Let  /Jb  be  the  measure  of  the  repulsive  force  at  a  unit's  dis- 
tance ;  then  the  forces  will  be 

,  and  mg  =  w  =  the  weight  of  the  particle. 


7»      2Ry 

'  2Ry    r'  2Ry'  r  '  '^Ry    r  ' 


which  in  (56)  give,  after  reduction, 

which  in  (5)  gives,  ^  =  -^  R, 

To  see  if  these  values  satisfy  equation  (57),  substitute  in  it 
the  values  of  X,  Z",  Z,  and  the  final  values  of  y  and  t*,  and 
we  find, 

xdx  4-  ydy  —  Rdy  +  zdz  =  0 ; 

which  is  the  differential  of  equation  (5),  and  hence  is  true. 
[This  is  the  theory  of  the  Electroscope.'] 

8.  A  particle  on  the  surface  of  an  ellipsoid  is  attracted  by 
forces  which  vary  directly  as  its  distance  from  the  principa 
planes  of  section ;  determine  the  position  of  equilibrium. 


[50,  51.] 

Let 


MOMENTS  OF  FORCES. 


<j  o  o 


79 


be  the  equation  of  the  surface 


fdL 


dL\  _  2^     (dL\  _  2v      fdL\ 


ill     f^IL\  -  ?f . 


and  let  the  x,  y,  and  z- components  of  the  forces  be  respec- 
tively, 

X=-fiiX,        Y=—fJii/,        Z=—fi^; 
and  (56)  will  give, 

which  simply  establishes  a  relation  between  the  constants ;  and 
hence  when  this  relation  exists  the  particle  may  be  at  rest  at 
any  point  on  the  surface. 

The  result  may  be  put  in  the  form, 


f^i 


/^i  +  /^  +  /Xg 


MOMENTS   OF   FORCES. 

60.  Def.  The  moment  of  a  force  in  reference  to  a  jpoini 
is  the  product  arising  from  m^ultijplying  the  force  hy  the 
perpendicular  distance  of  tlie  action-line  of  the  force  from 
the  point. 

Thus,  in  Fig.  28,  if  O  is  the  point 
from  which  the  perpendicular  is 
drawn,  F  the  force,  and  Oa  the 
perpendicular,  then  the  moment  of 
Fi% 

F.Oa  =  Ff', 

in  which  y  is  the  perpendicular  Oa. 

Tia.  98. 

61,  Nature  of  a  moment.  The  moment  of  a  force 
measures  the  turning  or  twisting  effect  of  a  force.  Thus,  in 
Fig.  28,  if  the  particle  upon  which  the  force  F  acts  is  at  Ay 


80  MEASURE  OF  A  MOMENT.  [5^-54.1 

and  if  we  conceive  that  the  point  0  is  rigidly  connected  to 
J.,  the  force  will  tend  to  move  the  particle  about  O^  and  it 
is  evident  that  this  effect  varies  directly  as  F.  If  the  action- 
line  of  F  passed  through  0  it  would  have  no  tendency  to 
move  the  particle  about  that  point,  and  the  greater  its  dis- 
tance from  that  point  the  greater  will  be  its  effect,  and  it  will 
vary  directly  as  that  distance ;  hence,  the  measure  of  the  effect 
of  a  moment  varies  as  thejproduct  of  the  force  and  jperjpendi- 
cular  ;  or  as 

oFf; 
where  <?  is  a  constant.     But  as  c  may  be  chosen  arbitrarily,  we 
make  it  equal  to  unity,  and  have  simply  Ff  as  given  above. 

52.  Def.  The  point  0  from  which  the  perpendiculars  are 
drawn  is  chosen  arbitrarily,  and  is  called  the  origin  of  mo- 
ments. When  the  system  is  referred  to  rectangular  coordinates, 
the  origin  of  moments  may,  or  may  not,  coincide  with  the 
origin  of  coordinates.  The  solution  of  many  problems  is 
simplified  by  taking  the  origin  of  moments  at  a  particular 
point. 

53.  The  lever  arm,  or,  simply,  the  arm^  of  a  force  is  the 
perpendicular  from  the  origin  of  moments  to  the  action-line 
of  the  force.  Thus,  in  Fig.  29,  Oa  is  the  arm  of  the  force  F^ ; 
Oc  that  of  the  force  i^ ,  etc.  Generally^  the  arm  is  the  per- 
pendicular distance  of  the  action-line  from  the  axis  about 
which  the  system  is  supposed  to  turn. 

54.  The  sign  of  a  moment  is  considered  "positive  if  it 
tends  to  turn  the  system  in  a  direction  opposite  to  that  of 
the  hands  of  a  watch ;  and  negative^  if  in  the  opposite  direc- 
tion. This  is  arbitrary,  and  the  opposite  directions  may  be 
chosen  with  equal  propriety  ;  but  this  agrees  with  the  direction 
in  which  the  angle  is  computed  in  plane  trigonometry.  Gen- 
erally we  shall  consider  those  moments  as  positive  which 
tend  to  turn  the  system  in  the  direction  indicated  by  the  nat- 
ural order  of  the  letters,  that  is,  ^positive  from  +  a;  to  •\r  y ; 
from  4-  y  to  -h  s ;  then  from  +  s  to  +  a? ;  and  negative  in  the 
reverse  direction. 

The  value  of  a  moment  may  be  represented  by  a  straight 
line  drawn  from  the  origin  and  along  the  line  about  which 


r55-59.] 


MOMENT  OF  A  FORCE. 


81 


rotation  tends  to  take  place,  in  one  direction  for  a  positive 
value,  and  in  the  opposite  direction  for  a  negative  one. 

65,  The  composition  and  resolution  of  moments  m.'i}^ 
be  effected  in  substantially  the  same  manner  as  for  forces. 
They  may  be  added,  or  subtracted,  or  compounded,  so  that  a 
resultant  moment  shall  produce  the  same  effect  as  any  num- 
ber of  single  moments.  The  general  proof  of  this  proposi- 
tion is  given  in  the  next  Chapter, 

56.  A  MOMENT  AXIS  is  a  line  passing  through  the  origin  of 
moments  and  perpendicular  to  the  plane  of  the  force  and 
arm. 

57.  The  moment  of  a  foeoe  refebred  to  a  moment  axis 
is  thejprodvAit  of  the  force  into  the  jperpendicular  distance  of 
the  force  from  the  axis. 

If,  in  Fig.  29,  a  line  is  dra^vn  through  O  perpendicular  to 
the  plane  of  the  force  and  arm,  it  will  be  a  moment  axis,  and 
the  turning  effect  of  i^  upon  that  axis  will  be  the  same  wher- 
ever applied,  providing  that  its  arm  Oa  remains  constant. 

If  the  force  is  not  perpendicular  to  the  arbitrarily  chosen 
axis,  it  may  be  resolved  into  two  forces,  one  of  which  will  be 
perpendicular  (but  need  not  intei-sect  it)  and  the  other  parallel 
to  the  axis.  The  moment  of  the  former  component  will  be 
the  same  as  that  given  above,  but  the  latter  will  have  no  mo- 
ment in  reference  to  that  axis  although  it  may  have  a  moment 
in  reference  to  another  axis  perpendicular  to  the  former. 

58.  The  moment  of  a  force  in  reference  to  a  plane 
to  which  it  is  parallel  is  the  product  of  the  force  into  the 
distance  of  its  action-line  from  the  plane. 

59.  If  obriy  numher  of  concurring  forces  are  in  equilibrium 
the  algebraic  sum  of  their  moments  will  be  zero. 

Let  i^,  7^,  7^,  etc..  Fig.  29,  be  the 
forces  acting  upon  a  particle  at  A ; 
and  O  the  assumed  origin  of  moments. 
Join  0  and  J.,  and  let  fall  the  per- 
pendiculars Oa,  Ob,  Oc,  etc.,  upon  the 
action -lines  of  the  respective  forces, 
and  let 

Oa=f\  Ob=f-,  Oc-f\  etc.  fio.  29. 


82 


BESULTANT  MOMENT. 


[60.1 


Resolve  the  forces  perpendicularly  to  the  line  OA;  and 
since  they  are  in  equilibrium,  the  algebraic  sum  of  these  com- 
ponents will  be  zero  ;  hence, 

F^  sin  OAF^  +i^  sin  OAF^,  +  F^  sin  OAF^  +  etc.  =  0 ; 

T^Oa       ^  Oh       ^  Oc 
or,i^^  +  i^^+i^s^  +  etc.  =  0. 

Multiply  by  OA^  and  we  have 

F^Oa  +  F^Oh  +  F^Oc  +  etc.  =  0; 
or,  i^/i  +  i^/2  +  i^/3  +  etc.  =  2Ff=  0.        (60) 
It  is  evident  that  any  one  of  these  moments  may  be  taken  as 
the  resultant  of  all  the  others. 

MOMENTS    OF    CONCURRING   FORCES    WHEN  THE  SYSTEM  IS    REFEREBD 

TO   RECTANGULAR   AXES. 

Y 

V 

60.    Let  A,  Fig.  30,   be   the 

point  of  application  of  the  forces 
i^,  i^,  i^,  etc.,  and  0  the  ori- 
gin of  coordinates,  and  also  the 
origin  of  moments.  Let  a?,  y, 
and  z  be  the  coordinates  of  the 
point  A.  Eesolving  the  forces 
parallel  to  the  coordinate  axes, 
we    have,  from    equation    (50), 


c 

A 

/ 

\ 

/ 

^    Y* 

B 

^ 

\4 

J) 

L 

/ 

/ 

Fzo. 


X 
Y 
Z 


sFgos  a; 
:^F(ios^; 
:s  Fcos  7. 


The  X-forces  prolonged  will  meet  \kiQjplane  of  yz\VLB\  and 
will  tend  to  turn  the  system  about  the  axis  of  y,  in  reference 
to  which  it  has  the  arm  BO=z\  and  also  about  z,  in  refer- 
ence to  which  it  has  the  arm  BD  =  y.  Hence,  employing 
the  notation  already  established,  we  have  for  the  moment  of 
the  sum  of  the  components  parallel  to  a?, 


—  Xy,  and  -f-  X2. 


[60.] 


MOMENTS  OF  CONCURRING  FORCES. 


88 


Similarly  for  the  y-components  we  find  the  moments, 

-f  Jaj,and—  Tz ; 

and  for  the  z-comjponents^ 

—  Za?,  and  +  Zy. 

The  moment  Xy  tends  to  turn  the  system  one  way  about  the 
axis  of  2,  and  Yx  tends  to  turn  it  about  the  same  axis,  but  in 
the  opposite  direction ;  and  hence,  the  combined  effect  of  the 
two  will  be  their  algebraic  sum  ;  or 

Yx  -  Xy. 

But  since  there  is  equilibrium  the  sum  will  be  zero.  Com- 
bining the  others  in  the  same  manner,  we  have,  for  the 
moments  of  concurring  forces,  in  equilibrium: 

In  reference  to  the  axis  of  aj  .  .  .  .  Zy  —  Js  =  0 ; 


y 


(61> 


.  .  Xz  —  Zc  =  0 ; 
"  "        "     "     "     "  2  .  .  .  .  rb  -  Xy=  0. 

The  third  equation  may  be  found  by  eliminating  z  from  the 
other  two ;  hence,  when  X,  Y,  and  Z  are  known,  they  are  the 
equations  of  a  straight  line ;  and  are  the  equations  of  the 
resultant. 

If  the  origin  of  moments  be  at  some  other  point,  whose 
coordinates  are  x',  y\  and  z' ;  and  the  coordinates  of  the  point 
A  in  reference  to  the  origin  of  moments  be  ajj,  yi,and  %; 
then  will  the  lever  arms  be 


a?!  =  a; 


2/i  =  y  —  y' ;  and  «i  =  3  —  s'. 


When  the  system  is  referred  to 
rectangular  coordinates  the  arm  of 
the  force,  referred  to  the  s-axis,  is 

xcoa^  —  y  cos  a, 

in  which  y  and  x  are  the  coordinates 

of  a7iy  point  of  the  action -line  of 

the  force ;  and  a  is  the  angle  which 

the  action-line  makes  with  the  axis  of  x,  and  y8  the  angle  which 

it  makes  with  y. 


Fio,  31. 


84 


EXAMPLES. 


[60.] 


In  Fig.  31,  let  BF  be  the  action-line  of  the  force  F^  O 
the  origin  of  coordinates,  A  any  point  in  the  line  AF^  cA 
which  the  coordinates  x  =  Ob,  and  y  =  Ah.  Draw  Od  and 
he  perpendicular  to  AF,  and  hd  from  h  parallel  to  BF.  The 
origin  of  moments  being  at  O,  Oa  v/ill  be  the  arm  of  the 
force. 
We  have 

Agh  =  a  =  chA, 
cAh  =  l3  =  hOd, 
ch  =  y  cos  a  =  ad, 
Od  =  X  cos  fi ; 
.*.  Oa  —  Od  —  ad  —  xco&fi  —  2/ cos  oc.        (61a) 

If  there  are  three  coordinate  axes,  this  will  be  the  arm  in 
reference  to  the  axis  of  s ;  and  if  there  be  many  forces,  the 
sum  of  their  moments  in  reference  to  that  axis,  will  be 
SF{x  (iosfi  —  y  cos  a). 

Examples. 

1.  A  weight  W  is  attached  to  a  string,  which  is 
secured  at  A,  Fig.  32,  and  is  pushed  from  a 
vertical  by  a  strut  OB ;  required  the  pressure 
Foil  BO  when  the  angle  OAB  is  6. 

The  forces  which  concur  at  B  are  the  weight 
TT,  the  pressure  F,  and  the  tension  of  the  string 
AB.  Take  the  origin  of  moments  at  A,  and 
we  have 

-  W.BO+  F.AO+  tension  x  0  =  0 ; 

/.i^=  TF-^=Ftanl9. 

2.  A  brace,  AB,  rests  against  a  vertical 
wall  and  upon  a  horizontal  plane,  and 
supports  a  weight  W  at  its  upper  dnd  ; 
required  the  compression  upon  the  brace 
and  the  thrust  at  A  when  the  angle  OAB 
is  6 ;  the  end  B  being  held  by  a  string  B  O. 

Fto.8& 


).] 


EXAMPLES. 


85 


The  concurring  forces  at  A^  are  TT,  acting  vertically  down- 
ward, the  reaction  of  the  wall  N  acting  horizontally,  and  the 
reaction  of  the  brace  F, 

Take  the  origin  of  moments  at  B^  we  have 
-  N,DB  +  W,CB  +F.  0  =  0; 
.•.jV  =  >f  tan^. 
Taking  the  origin  of  moments  at  D,  we  have 
W.  AD-F,  DB  sin  (9  +  iT.  0  =  0 ; 
.'.  ir=  TTseci?. 

3.  A  rod  whose  length  \^  BC=l  is  secured  at  a  point  B,  in 
a  horizontal  plane,  and  the  end  C  is  held  up  by  a  cord  AC 
so  that  the  angle  BA  Oia  6,  and  the  distance  AB  —  a ;  required 
the  tension  on  AG  and  compression  on  BC,  due  to  a  weight 

TF  applied  at  C, 

Ana.    Compression  =  —  TTcot  Q, 

4.  A  cord  whose  length  ABC  =  I  is  secured  at  two  points 
in  a  horizontal  line,  and  a  weight  W  is  suspended  from  it  at 
B ;  required  the  tension  on  each  part  of  the  cord. 


^jCo-jfi  -   ^ 


ty  ^  T^.a.d 


77^^=:^ 


CHAPTEE  in. 


PAEAIiLEL  FOBCES. 


61,  Bodies  are  extended  masses,  and  forces  may  be  applied 
at  any  or  all  of  their  points,  and  act  in  all  conceivable  direc- 
tions, as  in  Fig.  34. 


62,  StOPPOSE  THAT  THE  ACTION- 
LINES  OF  ALL  THE  FOECES  ARE  PARAL- 
LEL TO  EACH  OTHER.  This  is  a  Spe- 
cial case  of  concurrent  forces,  in 
which  the  point  of  meeting  of 
the  action-lines  is  at  an  infinite 
distance.  In  Fig.  35,  let  the  points 
a,  J,  G,  etc.,  which  are  on  the  action- 
lines  of  the  forces  and  within  the 
body,  be  the  points  of  applica- 
tion of  the  forces,  and  0  the  point 
where  they  would  meet  if  pro- 
longed. If  the  point  O  recedes 
from  the  body,  while  the  points 
of  application  a,  5,  c,  etc.  remain 
fixed,  the  action-lines  of  the  forces 
will  approach  parallelism  with 
each  other,  and  at  the  limit  will 
be  parallel. 


Fia.  84. 


^  0 


Fio.  85. 


63,  Resultant  of  parallel  forces.  The  forces  being  par- 
allel, the  angles  which  they  make  with  the  respective  axes, 
including  those  of  the  resultant,  will  be  equal  to  each  other 
Hence, 


[64.]  MOMENTS  OP  PARALLEL  FORCES.  87 

a  =  Oi  =  cUi  =  Ozi  6tc.  =  a ; 
5  =  A  =  A^/88,etc.  =  /Q; 
<J  =  7i  =  72  =  73  >  etc.  =  7  ; 
and  these,  in  equations  (50)  and  (51),  give 
X=  a  cosa  =  {Fi  +  Fi-h  Fs-^  etc.)  cos  a. 
F=  Ecosff  =  {Fi-rF2  +  Fs-h  etc.)  cos  y3.  (62) 

Z=/?cos7  =  (i^  +  i^  +  ii5+  etc.)  cos  7. 
From  either  of  these,  we  have 

^  =  i^  +  i^2  4-i^+  etc.  =  SF.  (63) 

Hence,  the  resultant  of  parallel  forces  eqicals  the  algebrwU 
sum  of  the  forces. 

From  (62),  we  have 

E=zVX^TY^TZ% 
which  is  the  same  as  (52). 

MOMENTS   OF   PARALLEL   FORCES. 

64.  Let  Fi,  F2,  Fz,  etc.,  be  the  forces,  and  Xi,  y^,  z^ ;  ojg,  yj, 
^2,  etc.,  be  the  coordinates  of  the  points  of  application  of  the 
forces  respectively  (which,  as  before  stated,  may  be  at  any 
point  on  their  action-lines).  Then  the  moments  of  F^  will  be, 
according  to  Article  (60), 

in  reference  to  the  axis  of  a?,  F^  cos  y^yi  —  Fi  cos  fi.Zi\ 
«  «        «    "      "     "  y,  i^Icosa.Si -i<;cos7.(Bi; 

"  "        "    "      "    "  ;3,  i^cos^.aji-i^icosa.yi; 

and  similarly  for  all  the  other  forces.     Hence,  the  sum  of  the 
moments  in  reference  to  the  respective  axes  for  equilibrium  is, 

{F{y^  +  i^2  +  ^3  +  etc.)  cos  7)      ^ 


-(^2i  +  ^  +  F^2^  +  etc.)  cos  /9 ) 

iF^^  +  i^  +  ife  +  etc.)  cos/9) 
-(^i  +  ^8  +  ^'Jys  +  etc.)  cos  a ) 


(75^01  4-  ifej  +  ife  +  etc.)  cos  a 
— (i^iCi  -f  ife  +  FiPi^  +  etc.)  cos  7 


88  THBEB  PARALLEL  FORCES.  [66.] 

These  equations  will  be  true  for  all  values  of  a,  )3,  and  7,  if 
the  coefficients  of  cos  a,  cos  ^8,  cos  7,  are  respectively  equal  to 
zero ;  for  which  case  we  have 


F^x^  +  i^  +  F^x^  +  etc.  =  ^Fx  =  0 ; 
F^yi  4-  F^y^  +  F^y,  +  etc.  =  SFy  =  0; 
F,2,  +  F^s.^  4-  i^%  +  etc.  =  SF2  =  0 ; 


(64) 


from  which  the  coordinates  of  the  point  of  application  of  any 
one  of  the  forces,  as  i^,  for  instance,  may  be  found  so  as  to 
satisfy  these  equations,  when  all  the  other  quantities  are  given. 
Let  the  given  forces  have  a  resultant.  Let  a  force,  as  Fi, 
equation  (64),  equal  and  opposite  to  the  resultant,  be  introduced 
into  the  system,  then  will  tliere  be  equilibrium.  Let  SFx,  SF/, 
SF3,  include  the  sum  of  the  respective  products  for  all  the 
forces  except  that  of  the  resultant ;  li  be  the  resultant,  and  x, 
y,  5,  the  coordinates  of  the  point  of  application  of  the  result- 
ant ;  so  chosen  as  to  satisfy  equations  (64),  then  we  have 

J2x-SFx=0;    By-SFy  =  0;    RJ-XFz^^,    (65) 

Substitute  the  value  of  ^  =  XF^  in  these  equations,  and  we 
find 

_      XFx^  -      XFy^  _      XFz^ 

^=XP'  y=-%F'  ^^'XF'  ^^^) 

by  which  the  point  of  application  of  the  resultant  becomes 
known,  and,  being  independent  of  a,  yS,  and  7,  is  a  jpoint 
through  which  the  resultant  constantly  passes^  as  the  forces  a>re 
turned  about  their  jpoints  of  application^  the  forces  constantly 
retaining  their  parallelism.  This  point  is  called  the  centre  of 
parallel  forces. 

65.  If  the  system  consists  of  three  forces  only,  and  are 
in  tlie  plane  icy,  we  have 

R  =  F,^-F^', 
FW^rF^,  _  F^,  +  Fiy,      ).         (67) 

^-~^r+X~'  ^~     F,  +  F,' 


[66.J 


STATICAL  COUPLES 
Y 


let.  Consider  F^  arid  Fz  as  positive. 
The  resultant  will  equal  the  arith- 
metical sum  of  the  forces.  Take 
the  origin  at  a  Fig.  36,  where  the 
resultant  cuts  the  axis  of  x;  then 
X  =  0,  and  the  second  of  (67)  gives 

Fix,=  -F^] 
and  hence,  if  i^  >  i^,  a?2  will  exceed 
Xi ;  that  is,  the  resultant  is  nearer  the 
greater  force. 

2d.  Consider  F^  at  negative. 

In  this  case  the  resultant  equals  thj 
difference  of  the  forces.  Take  the  origin 
at  a,  Fig.  37,  and  we  have 

F^x^  =  F^ ; 

and  hence  both  forces  are  either  at  the 
riofht  or  left  of  the  resultant. 


r. 


F. 


Pio.  86. 


F. 


r, 


Fio.  87. 


3d.  Let  Fx  =  Fi^  F^  and  one  of  the  forces  be  negative,  then 


R  =  F-F=Oix  = 


-      F{x,±x^) 


F-F 


—  -o  ;  and  y  =  co  ;    {foS) 


that  is,  the  resultant  is  zero,  while  the  forces  may  have  a  finite 
moment  equal  to  Fi^Xi  ±  x^.     Such  systems  are  called 


COUPUES. 

66.  ^  Goujple  consists  of  two  eq^iial  parallel  forces  acting  in 
opposite  directions  at  a  finite  distance  from  each  other, 

A  statical  couple  cannot  be  equilibrated  by  a  single  force. 
It  does  not  produce  translation,  but  simply  rotation.  A  couple 
can  he  equilibrated  only  by  an  equivalent  couple. 

Equivalent  couples  are  such  as  have  equal  moments. 

The  resultant  of  several  couples  is  a  single  couple  which 
will  produce  the  same  effect  as  the  component  couples. 


Y 


90  AXIS  OF  A  COUPLE.  [67-68.J 

67.  The  arm  of  a  couple  is  the  jperpendicular  distaytct 
hetween  the  action-lines  of  the  forces. 
Thus,  in  Fig.  38,  let  O  be  the  origin 
of  coordinates,  and  the  axis  of  x  per- 
pendicular to  the  action-line  of  F'^ 
—  X  then  will  the  moment  of  one  force  be 
Fxi ,  and  of  the  other  Fx*^ ,  and  hence 
the  resultant  moment  will  he 


X,z 


F 


Xx 


Fio. 


F{x^-x^)  =  F.al',  (69) 


hence,  db  is  the  arm.  If  the  origin  of  coordinates  were 
between  the  forces  the  moments  would  he  F{xi-\-  x^  =  F.db 
as  before.  If  the  origin  be  at  a  we  have  FS)  -j-  F.ab  —  F.ab  as 
before. 

68-  The  axis  of  a  statical  couple  is  any  line  jperpen- 
dicular to  the  plane  of  the  couple.  The  length  of  the  axis 
may  be  made  proportional  to  the  moment  of  the  couple,  and 
placed  on  one  side  of  the  plane  when  the  moment  is  positive, 
and  on  the  opposite  side  when  it  is  negative ;  and  thus  com- 
pletely represent  the  couple*  in  magnitude  and  direction. 

If  couples  are  in  parallel  planes^  their  axes  may  be  so  taken 
that  they  will  conspire,  and  hence  the  resultant  couple  equals 
the  algebraic  sum  of  all  the  couples. 

If  the  planes  of  the  couples  intersect,  their  axes  may 
intersect. 

Let  O  =  F.ah  =  the  moment  of  one  couple ; 

Oi=  F^.aJ)i  =  the  moment  of  another  couple ; 
0  —  the  angle  between  their  axes ;  and 
On  =  the  resultant  of  the  two  couples ; 


and  this  resultant  may  be  combined  with  another  and  eo  on 
until  the  final  resultant  is  obtained . 

Examples. 
1.  Three  forces  represented   in   magnitude,  direction  and 
position,  by  the  sides  of  a  triangle,  taken  in  their  order,  produce 
a  couple. 


[68.] 


EXAMPLES. 


91 


Fio. 


2.  If  three  forces  are  represented  in  magnitude  and  position 
by  the  sides  of  a  triangle,  but  whose  directions  do  not  follow  the 
order  of  the  sides  ;  show  that  they  will  have  a  single  resultant. 

3.  On  a  straight  rod  are  suspended  several  weights ;  F^  — 

5  lbs.,  i^  =  15  lbs.,  i^  =  7  lbs., 
F^^^  lbs.,  i^  =  9  lbs.,  at  dis 
tances  AB  =  3  ft.,  BD  =  6  ft., 
I>F=  5  ft.,  and  FF  =  A  ft.  ; 
required  the  distance  AC  B.t 
which  a  fulcrum  must  be  placed 
80  that  the  weights  will  balance 
on  it ;  also  required  the  pressure  upon  it. 

4.  The  whole  length  of  the  beam  of  a  false  balance  is  2  feet 

6  inches.  A  body  placed  in  one  scale  balances  6  lbs.  in  the 
other,  but  when  placed  in  the  other  scale  it  balances  8  lbs. ; 
required  the  true  weight  of  the  body,  and  the  lengths  of  the 
arms  of  the  balance. 

5.  A  triangle  in  the  horizontal  plane  x,  y  has  weights  at  the 
sjveial  angles  which  are  proportional  respectively  to  the 
opposite  sides  of  the  triangle ;  required  the  coordinates  of  the 
centre  of  the  forces. 

Let  a?! ,  ^1  be  the  coordinates  of  A^ 
Xijy^oiB;  a^,ysof  C; 
X,  y  of  the  point  of  application  of  the  ro^.^ltant ; 
then  we  have 

(a  +  J  +  c)fiB=aaJi  +  Ja!i,  +  ca%;  and 
\a-\-l  -^  c)y  =ay^-\-  hy^  +  cy^. 

6.  If  weights  in  the  proportion  of  1,  2,  3,  4,  5,  6,  7  and  8  are 
suspended  from  the  respective 
angles  of  a  parallelopiped ;  re- 
quired the  point  of  application  .  Fi 
of  the  resultant. 

7.  Several  couples  in  a  plane, 
whose   forces   arc   parallel,  are 
applied  to  a  rigid  right  line,  as 
in   Fig.  40  ;   required    the  re-  •** 
Bultant  couple.  ^^Jj,^ 


F, 


Fj 


CENTRE  OF  GRAVITY 


M 


8.  Several  couples  in  a 
plane,  whose  respective 
arms  are  not  parallel,  as 
in  Fig.  41,  act  upon  a  rigid 
right  line ;  required  the 
resultant  couple. 


CENTRE   OF   GBAVITY   OF   BODIES. 

69.  The  action-lines  of  the  force  of  gravity  are  normal  to 
the  surface  of  the  earth,  but,  for  those  bodies  which  we  shall 
here  consider,  their  convergence  will  be  so  small,  that  we 
may  consider  them  as  parallel.  We  may  also  consider  the 
force  as  the  same  at  all  points  of  the  body. 

The  centre  of  gravity  of  a  body  is  the  point  of  application 
of  the  resultant  of  the  force  of  gravity  as  it  acts  upon  every 
particle  of  the  body.  It  is  the  centre  of  parallel  forces.  If 
this  point  be  supported  the  body  will  be  supported,  and  if  the 
body  be  turned  about  this  point  it  will  remain  constantly  in 
the  centre  of  the  parallel  forces. 

Let      M  =  the  mass  of  a  body  ; 

m  =  the  mass  of  an  infinitesimal  element ; 

V  =  the  volume  of  the  body ; 

D  =  the  density  at   the  point  whose  coordinates  are 

X,  y,  and  z ; 
_R  1=  W  =  the  resultant  of  gravity,  which  is  the  weight ; 

and 
X,  y,  and  i  be  the  coordinates  of  the  centre  of  gravity. 

We  have,  according  to  equations  (63)  and  (20), 
a  =  W=  Sgm  =  M  X  g; 
and  (65)  becomes 

X  Sgm  =  Sgmx ;  or  Mx  =  Smx ; 
y  ^gm  =  ^gmy ;  or  My  =  ^?/iy 
z  'Sgrn  =  Sgms ;  or  Ms  =  Sms. 

If  the  density  is  a  continuous  function  of  the  coordinates  of 
the  body  we  may  integrate  the  preceding  expressions.      The 


n 


m 


OF  BODIES. 


93 


complete  solution  will  sometimes  require  two  or  three  integra- 
tions, depending  upon  the  character  of  the  problem  j  but,  using 
only  one  integral  sign,  (22)  and  (70)  become 


^/DdV^fDxdV', 
y/DdV^/DydV', 
zfDdV^fDzdY, 


(71) 


If  the  origin   of   coordinates  be  at  the  centre  of  gravity, 
then 


aj  =  0;  y  —  0;  s  =  0; 
and  hence, 

Xmx=J'DxdY=^\ 
and  similarly  for  the  other  values. 
If  D  be  constant,  this  becomes 


(71a) 


(71J) 


the  limits  of  integration  including  the  whole  body. 

If  the  mass  is  homogeneous,  the  density  is  uniform,  and  D 
being  cancelled  in  the  preceding  equations,  we  have 


-       fxdV 

if  TT 


-        /zdV 


(72) 


Many  solutions  may  be  simplified  by  observing  the  following 
principles : 

1.  If  the  body  has  an  axis  of  8ym>metry  the  centre  of  gravity 
will  he  on  that  axis* 


94  CENTRE  OP  GRAVITY  [70.1 

2.  If  the  hodyhas  ajplane  of  symmetry  the  centre  of  gravity 
will  he  in  that  jplane. 

3.  If  the  hody  has  two  or  more  axes  of  symmetry  the  centre 
of  gravity  will  he  at  their  intersection. 

Hence,  the  centre  of  gravity  of  a  physical  straight  line  of 
uniform  density  will  be  at  the  middle  of  its  lengtli ;  that  of 
the  circumference  of  a  circle  at  the  centre  of  the  circle  ;  that 
of  the  circumference  of  an  ellipse  at  the  centre  of  the  ellipse  ; 
of  the  area  of  a  circle,  of  the  area  of  an  ellipse,  of  a  regular 
polygon,  at  the  geometrical  centre  of  the  figures.  Similarly 
the  centre  of  gravity  of  a  triangle  will  be  in  the  line  joining 
the  vertex  with  the  centre  of  gravity  of  the  base ;  of  a  pyramid 
or  cone  in  the  line  joining  the  apex  with  the  centre  of  gravity 
of  the  base. 

There  is  a  certain  inconsistency  in  speaking  of  the  centre  of 
gravity  of  geometrical  lines,  surfaces,  and  volumes  ;  and  when 
they  are  used,  it  should  be  understood  that  a  line  is  iijphysical 
or  material  line  whose  section  may  be  infinitesimal ;  a  surface 
is  a  material  section^  or  thin  plate,  or  thin  shell ;  and  a  volume 
is  a  mass,  however  attenuated  it  may  be. 

When  a  body  has  an  axis  of  symmetry,  the  axis  of  x  may  be 
made  to  coincide  with  it,  and  only  the  first  of  the  preceding 
equations  will  be  necessary.  If  it  has  a  plane  of  symmetry, 
the  plane  x  y  may  be  made  to  coincide  with  it,  and  only  the 
first  and  second  will  be  necessary. 

70.  Centre  of  gravity  of  material  lines. 

Let  h  =  the  transverse  section  of  the  line,  and 
ds  =  an  element  of  the  length, 
then 

dV=kds; 

and  (71)  becomes 

^/mds  =  fDhcds ; 

yfBMs  =fDkyds;  [  (73) 

'^fD^ds  =  fDkzds. 


)70.1 


OF  LINES. 


^  8 

-  _  y^^ 

~"         8 


(74) 


95 


If  the  transverse  section  and  the  density  are  uniform,  we 
have 


The  centre  of  gravity  will  sometimes  be  outside  of  the  line 
or  body,  and  hence,  if  it  is  to  be  supported  at  that  point,  we 
must  conceive  it  to  be  rigidly  connected  with  the  body  by 
lines  which  are  without  weight. 


Examples.       • 

1.  Find  the  centre  of  gravity  of  a  straight  fine  wire  of  uni- 
form section  in  which  the  density  varies  directly  as  the  distance 
from  one  end. 

Let  the  axis  of  x  coincide  with  the  line,  and  the  origin  be 
taken  at  the  end  where  the  density  is  zero.  Let  8  be  the  density 
at  the  point  where  a?  =  1 ;  then  for  any  other  point  it  will  be 
D  =  Sx;  and  substituting  in  the  first  of  (73),  also  making 
d8  =  dxy  we  have 


w  I  Bxdx  —    /  Ba^dx; 
^0  ^0 


ia. 


This  corresponds  with  the  distance  of  the  centre  of  gravity 
of  a  triangle  from  the  vertex. 


Fia.  43.  Tio.  43. 

S.  Find  tlie  centre  of  gravity  of  a  cone  or  pyramid,  whether 


90  EXAMPLES.  [70.1 

right  or  oblique,  and  whether  the  base  be   regular  or  irre- 
gular. 

Draw  a  line  from  the  apex  to  the  centre  of  gravity  of  tiie 
base,  and  conceive  that  all  sections  parallel  to  the  base  are  re- 
duced to  this  central  line.  The  problem  is  then  reduced  to 
finding  the  centre  of  gravity  of  a  physical  line  in  which  the 
density  increases  as  the  square  of  the  distance  from  one  end. 

Ans.  X  =  ia, 

3.  To  find  the  centre  of  gravity  of  a 
circular  arc. 

Let  the  axis  of  x  pass  through  the 
centre  of  the  arc,  B.  and  the  centre 
of  the  circle  O ;  then  Y=0.  Take 
the  origin  at  B : 

and  let  cc  =  BD, 


25  =  the  arc  A  BC,  and 
r  =  OC  =  the  radius  of  the  circle ; 


then, 


f  =  2rx  -  ar». 

Differentiate,  and  we  have 

ydj/  =  rdx  —  xdx ; 

dy        dx    .  .  .        ds 
=  —  which  =  — : 


r—x        y  r 

hence,  the  first  of  (74)  gives 

^        xdx 


p 

Jo  V2rx-x'       rf      ^r^ -.        1^  ry 

=:  — I  —  V  2rx  —  ar  +  5  I    =  r -: 

^  L  Jo       .  ^ 


_  V  2rx  —  ^ 

X  =  . 


which  is  the  distance  Be.    Then  cO  —  r  —  ir  —  — ^=  — > 
hence,  the  distance  of  the  centre  of  gravity  of  an  arc  from  the 


[71.]  CENTRE  OF  GRAVITY  OF  SURFACES.  97 

centre  of  the  circle  is  a  fourth  proportional  to  the  arc,  the 
radius,  and  the  chord  of  the  arc. 

71.  Centre  of  gravity  of  plane  surfaces. 

Let  the  coordinate  plane  xy  coincide  with  the  surface  ;  then 
dV=dxdy\  .*.  V  =  ffdxdy  =  fydx  ov  fxdy\  and  (71)  be- 
comes 

xffndxdy=ffl>xdxdy;    ) 

yffBdxdy=ffDydxdy.     )  ^    ^ 

The  integrals  are  definite,  including  the  whole  area.  The 
order  of  integration  is  immaterial,  but  after  the  first  integra- 
tion the  limits  must  be  determined  from  the  conditions  of  the 
problem.  If  D  is  constant  and  the  integral  is  made  in  respect 
to  y,  we  have 

-^fyxdx         1 

!"  Cr6) 

-  _  Ufch  I 

y-  fydx'^      ] 

and  if  x  be  an  axis  of  synimetry,  the  first  of  these  equations 
will  be  sufficient. 

If  the  surface  is  referred  to  polar  coordinates,  then  <iF  = 
pdpdO,  and  x  =  p  cos  6,  y  =  p  sin  6,  and  (71)  becomes 

_^  ffDp^cosOdpde 
^  -       JfJJpdpdO       ' 


-_  ffDp'^^riedpde 
^  -     ffDpdpde 


(77) 


Examples. 
1.  Find  the  centre  of  gravity  of  a   semi-parabola  whose 
equation  is  y*  =  ^jpx. 

Equations  (76)  become 


ly^p 


X  dx 


y 


=  l»; 


i  J^  "ipxdx 


V  = 
7 


1^ 


;X  ^Z 


98  CENTRE  OF  GRAVITY.  [72.] 

2.  Find  the  centre  of  gravity  of  a  quadrant  of  a  circle  in 
whicli  the  density  increases  directly  as  the  distance  from  the 
centre. 

Let  8  =  the  density  at  a  unit's  distance  from  the  centre ; 
then 

I)  =  Bp  Sit  B.  distance  p ; 
and  (77)  becomes 


B  I    p^cosddpdO 

Vo    Jo  '''"'' 


^  =  —^rxJ  r^ ^^^'^^ 


72,  Centre  of  gravity  of  curved  surfaces. 

We  have  for  an  element  of  the  area 
dY-=^dx  dy  x  sec.  of  the  angle  hetween  the  tangent  joHane  and 
the  coordinate  j^lane  xy  ;  or, 

d  V=sec.  6xdx  dy; 
or,  in  terms  of  partial  differential  coefficients 


// 


./ ldL\'     (dLV     (dL\^ 
\dz) 


and,  for  a  homogeneous  surface,  (72)  becomes 

-     SMV. 

X  = y—  ) 

y        V 
v~' . 

or,  the  surface  may  be  referred  to  polar  coordinatea^ 


(78) 


[72.]  OP  DOUBLE  CURVED  SUEFACES.  99 

If  the  surface  is  one  of  revolution,  let  x  coincide  with  the 
axis  of  revolution,  then 

x/iryds  ^  firyxds,  (78a) 

Example. 

Find  the  centre  of  gravity  of  one-eighth  of  the  surface  of  a 
sphere  contained  within  three  principal  planes. 
Let  the  equation  of  the  sphere  be 

then 

and  the  first  of  (78)  becomes 


"^= fn? ^^y 

2     pp        X  dx  dy 


irR^^    VE'-f-a? 


X  =  VW^^ 


=  -  Afvi^-y'-^M 


=  :^fv{i^-^^ 


wR 

=  iR. 
Similarly,  y  =  ^R  =  s. 

This  problem  may  be  easily  solved  by  the  aid  of  elementary  geometry, 
Conceive  that  the  surface  is  divided  into  an  indefinite  number  of  small  zones 
by  equidistant  planes  which  are  perpendicular  to  the  axis  of  x^  in  which  case 


100 


CENTRE  OF  GRAVITY  OF  VOLUMES. 


[73.] 


the  area  of  the  zones  will  be  equal  to  each  other.  Conceive  that  these  zones 
are  reduced  to  the  axis  of  x\  they  will  then  be  uniformly  distributed  along 
that  axis,  and  hence  the  centre  of  gravity  will  be  iM  from  the  centre ;  and  aa 
the  surface  is  symmetrical  in  respect  to  each  of  the  three  axes,  we  get  the 
same  result  in  respect  to  each. 

73,  Centre  of  gravity  of  volumes^  or  heavy  todies. 
We  have 

dV  =  dxdydz\ 
/.  xfffD dxdydz^fff  Dxdxdydz\  (79) 

and  similarly  for  y  and  i. 

If  X  is  an  axis  of  symmetry^  (79)  is  sufficient. 

If  the  surface  is  referred  to  polar  coordinates,  let 

r 

^  =  AOx, 

e  =  dOA, 

P=  Og, 
then, 

gd  =  dp, 
gf=  pde, 

gh  =  p  COS  6d(j>y 

and, 

dV  =  p^dp  cos  6d6d4>, 

also, 
i         X  =  p  cos  6  cos  (jt;  y  =  p  sin  6 ;  and  z  =  p  cos  6  sin  <f>. 

Hence,  for  a  homogeneous  body,  we  have 

Vx  =fffp^cos^ecos(t>dpded<t>;^ 

Vy  ^Sff  P^cosS  sinddpded<l>\  \  (80) 

Yz^fffp^co^esin^dpdOd^.  j 

If  the  volume  be  one  of  revolution  about  the  axis  of  a?,  we 
have 


Fio.  45. 


[73.]  EXAMPLES.  101 

dY=z7tfdx;       \  ,    . 

Yx=7tffxdx.)  ^    ^ 

Example. 

Find  the  centre  of  gravity  of  one-eighth  of  the  volume  of  a 
homogeneous  ellipsoid,  contained  within  the  three  principal 
planes. 

Let  the  equation  of  the  ellipsoid  be 

then,  equation  (79)  gives, 

ro   pY    nX  re    rY    r*X 

so  I      j        I  dxdyd2=   I      I         I  xdxdydz. 

Jo  Jo    Jo  Jo  Jo    Jo 

Performing  the  integration,  we  have 

/.  uj  =  f  «. 
Similarly,  y  =  |5,  and  z  =  |c. 

Performing  the  above  integration  in  the  order  of  the  letters  a?,  y  and  «, 
and  using  the  limits  in  the  reverse  order  as  indicated,  we  have  for  the 
x-ltmit8. 


aj  =  0,anda.  =  a|/l_^-^  =  X, 
and  the  corresponding  limits  for  y  will  be 

y  =  b^l-^  =  r;  andy  =  0. 
For  the^r**  member  of  the  equation,  we  have 

Consider  \  /  1 ,=  B,  a  constant   in    reference   to  the  y-integraUon 

and  we  have 


4 

102  EXAMPLES.  [74.1 


For  the  second  member,  we  have 


(b. 


Consider  «  as  constant  in  performing  the  y -integration^  and  we  have 

-      ■^va'^hG       ,  ■      • 

as  given  above. 

74,  When  the  centre  of  gravity  of  a  body  is  known  and  the 
centre  of  gravity  of  a  part  is  also  known,  the  centre  of  the 
remaining  part  may  be  found  as  follows : — 

Let    1^=  the  weight  of  the  whole  body  ; 

X  =  the  distance  from  the  origin  to  the  centre  of  gravity 
of  the  body ; 
tOi  =  the  weight  of  one  part ; 
aJi  =  the  distance  of  the  centre  of  ^i  from  the  same 

origin  ; 
w^  =  the  weight  of  the  other  part ;  and 
axj  r=  the  distance  of  the  centre  of  i^^g  from  the  same 
origin ; 
then 

w^  +  Wc^  =  {wy^  -\-  W2)x=  Wx ; 
and  hence, 


'''-  i,'^'\(^^''^)'!'^ 


H" 


m.] 


THEOREMS  OF  PAPPUS. 


103 


If  the  body  is  homogeneous,  the  volumes  may  be  substituted 
for  the  weight. 

Example. 

Let  ABC  be  a  cone  in  which  the  line 
BE  joins  the  vertex   and  the  centre  of 
gravity  of  the  base;  and  the  cone  ADG^ 
having  its-  apex  Z>,  on  the  line  BE^  and 
the  same  base  ^  6^  be  taken  from  the  for- 
mer cone,  required  the  centre  of  gravity 
of  the  remaining  part,  ADCB, 
Let  F  =  the  volume  of  ACB, 
a  =  BE, 
<h  =  BE, 
X  =  Ec  —  ia  —  the  distance  of  the  centre  of  ABG 

from  E, 
Xi=:  Eb  =  iai=  the  distance  of  the  centre  of  ABC 
from  E; 
then, 

—  F  =  the  volume  of  ACB, 


Fio.  46. 


and  (82)  becomes 


<h 


X2  = 


F.  i  «  -  F.  ^.  i(h 
a 

F-  F^ 
a 


=  i{a  +  «i). 


Centkobario  Method,  ob 

75.  Theorems  of  Pappus  or  of  GuLDnsrus. 

Multiply  both  members  of  the  second  of  (74)  by  27r,  and  it 
may  be  reduced  to 

Qirys  =  ^Trfyds,  (83) 

the  second  member  of  which  is  the  area  generated  by  the  revo- 
lution of  a  line  whose  length  is  8  about  the  axis  of  x,  and  the 


104  EXAMPLES.  [75.] 

first  member  is  the  circumference  described  by  the  centre  of 
gravity  of  the  line,  multiplied  by  the  length  of  the  line  ;  hence, 
the  area  generated  hy  the  revolution  of  a  line  ahoiit  a  fixed 
axis  eqxials  the  length  of  the  line  multiplied  hy  the  circu  nfer- 
ence  described  hy  the  centre  of  gravity  of  the  line. 

This  is  one  of  the  theorems,  and  the  following  is  the  other. 

From  the  second  of  (76),  we  find 

27ry  A  —firy^dx. 

The  right-hand  member,  integrated  between  the  proper 
limits,  is  the  volume  generated  by  the  revolution  of  a  plane 
area  about  the  axis  of  x.  The  plane  area  must  lie  wholly  on 
one  side  of  the  axis.  In  the  first  member  of  the  equation,  A 
is  the  area  of  the  plane  curve,  and  27ry  is  the  circumference 
described  by  its  centre  of  gravity.  Hence,  the  volume  gener- 
ated hy  the  revolution  of  a  jplane  curve  which  lies  wholly  on 
one  side  of  the  axis,  equals  the  area  of  the  curve  multiplied  hy 
the  circumference  described  by  its  centre  of  gravity. 

Examples. 

1.  Find  the  surface  of  a  ring  generated  by  the  revolution  of 
a  circle,  whose  radius  is  a  about  an  axis  whose  distance  from 
the  centre  is  c. 

Ans.  4^  re. 

2.  The  surface  of  a  sphere  is  ^ttt^,  and  the  length  of  a 
semicircumference  isTrr;  required  the  ordinate  to  the  centre 
of  gravity  of  the  arc  of  a  semicircle. 

3.  Required  the  volume  generated  by  an  ellipse,  whose  semi- 
axes  are  a  and  J,  about  an  axis  of  revolution  whose  distance 
from  the  centre  is  c ;  c  being  greater  than  a  or  h. 

(Observe  that  the  volume  will  be  the  same  for  all  positions  < 
of  the  axes  a  and  h  in  reference  to  the  axis  of  revolution.) 

4.  The  volume  of  a  sphere  is  ^ir  7^,  and  the  area  of  a  semi- 
circle is  ^TT  7^ ;  show  that  the  ordinate  to  the  centre  of  gravitv 

.    —        4-r 
of  the  semicircle  is  y   =  ^. 


f76.J  EXA.MPLES.  105 

76»  Additional  Examples. 

1.  Fiiid  the  centre  of  gravity  of  the   quadrant  of  the  circnraference  of  a 
circle  contained  between  the  axes  x  and  ^,  the  origin  b<sing  at  the  centre. 

.        -       2r      - 
Ans.  X  =  —  =  y. 

2.  Find  the  distance  of  the  centre  of  gravitj  of  the  arc  of  a  cycloid  from  the 
rertex,  r  being  the  radius  of  the  generating  circle.  0  '       'f   \  { 

Ans.  y  =  |r. 

3.  Find  the  centre  of  gravity  of  one-half  of  the  loop  of  a  leminiscate,  of 
which  the  equation  is  r'  =  a'  cos  2d,  I  being  the  length  of  the  half  loop. 

a^      _       a»    Vj  _  1 

4.  Find  the  centre  of  gravity  of  the  helix  whose  equations  are 

X  =  a  cos  (p,  y  =  a  sin  <p,  3  =  na  <Pt 
the  helix  starting  on  the  axis  of  x, 

Ans.  x=na-\y=na  ;   and  z  =  \z, 

z  z 

5.  Find  the  centre  of  gravity  of  the  perimeter  of  a  triangle  in  space. 

6.  If  Xq  and  y^  are  initial  points  of  a  curve,   find  the  curve  such  that 
mx  =  X  —  Xo,  and  ny  =  y  —  yo» 

7.  A  curve  of  given  length  joins  two  fixed  points  ;  required  its  form  so  that 
its  centre  of  gravity  shall  be  the  lowest  possible. 

(This  may  be  solved  by  the  Calculus  of  "Variations). 

Ans.  A  Catenary. 

8.  Find  the  centre  of  gravity  of  a  trapezoid. 

Let  ADEB  be  the  trapezoid,  in  vrhich  DE  and  ,\^ 

AB  are  the  parallel  sides.     Produce  AD  and  BE  /  j  \ 

until  they  meet  in  C,  and  join  C  with  E,  the  mid-  /    /    \ 

die  point  of  the  base  ;  then  the  centre  of  gravity  will 
be  at  some  point  g  on  this  line.  The  centre  of  the 
triangle  A  CB  will  be  on  CE,  and  at  a  distance  of 
^CE  from  E;  and  simQarly  that  of  DCE  wUl  be 
on  the  same  line,  and  at  a  distance  of  i  CG  from 
O ;  then,  by  (82)  we  may  find  Eg. 

(If  DE  is  zero,  we  have  Eg  —  \EC  for  the  centre  of  gravity  of  the  triangle 
ABC.) 

9.  Find  the  centre  of  gravity  of  the  quadrant  of  an  ellipse,  whose  equation 
it  cV  4-  y«»  =  a«ft«. 

.        -         a      -      .  b 
Ans.   X  =^-;  y  =  |  — 


P  ■     ^    \e 

A  i 

/  k\ 

F 

Fio.  47. 

—  1 

^^AB  ^2I)E 

106 


EXAMPLES. 


[76.1 


10.  Find  the  centre  of  gravity  of  the  circular 
sector  ABC. 


Let  the  angle  ACS  =26;  then 


<B=Cg  =  l 


rfone 


11.  Find  the  centre  of  gravity  of  a  part  of  a 
circular  annulus  ABED, 
Let  AC  =  r,  JDG  =  r,,  and  ACB  =26 ;  then 
sin  6    r^  +  TTi  +  Ti* 


Ans.  Cg7  =  f 


r+Ti 


12.  Find  the  centre  of  gravity  of  the  circular  spandril  FQB. 

13.  Find  the  centre  of  gravity  of  a  circular  segment. 

(chord)* 


Am.  Diet,  from  G  = 


12  area  of  segment 


14.  Find  the  distance  of  the  centre  of  gravity  of  a  complete  cycloid  froir 
its  vertex,  r  being  the  radius  of  the  generating  circle. 

Ans.  y  =  Ir. 

15.  Find  the  centre  of  gravity  of  the  parabolic 
spandril  OCB,  Fig.  49,  in  which  OG  =  y,  and  GB 
=  x. 

Ans.  x  =  -^x\ 

16.  Find  the  centre  of  gravity  of  a  loop  of  the 
leminiscate,  whose  equation  is  r'  =  a'^  cos  26. 


Fia.  49. 


Ans.  x  =  —-a. 
2t 


17.  Find  the  centre  of  gravity  of  a  hemispherical  surface. 

Ans.  x  =  \r. 

18.  Find  the  centre  of  gravity  of  the  surface  generated  by  the  revolution  of 
a  semi-cycloid  about  its  base,  a  being  the  radius  of  the  generating  circle  and  2 
the  distance  from  the  vertex  of  the  surface. 

Ans.  X  =  rl^. 

19.  The  centre  of  gravity  of  the  vv>lume  of  a  paraboloid  of  revolution  ia 

«  =  §-«. 

20.  The  centre  of  gravity  of  one  half  of  an  ellipsoid  of  revolution,  of  wjtich 
the  equation  is  aV  +  b^x^  =  a'ft",  is 

5  =  |a. 

21.  The  centre  of  gravity  of  a  rectangular  wedge  is 


1 1   ''L 


[78.]  EXAMPLES.  107 

22.  The  centre  of  gravity  of  a  Bemicircular  cylindrical  wedge,  whose  radius 
isr,  ia 


Fio.  60.  Pio.  61. 

23.  The  vertex  of  a  right  circular  cone  is  ia  the  surface  of  a  sphere,  the 
axis  of  the  cone  passing  through  the  centre  of  the  sphere,  the  base  of  the  cone 
being  a  portion  of  the  surface  of  the  sphere.  If  2d  be  the  vertical  angle  of  the 
cone,  required  the  distance  of  the  centre  of  gravity  from  the  apex. 

.  1  —  C08*d 

1  —  cos*  6 

24.  Find  the  distance  from  (?,  Fig.  48,  to  the  centre  of  gravity  of  a  spheri- 
cal sector  generated  by  the  revolution  of  a  circular  sector  O  CA,  about  the 
axis  OC. 

Ans.  i(Qa+iQR). 

25.  A  circular  hole  with  a  radius  r  is  cut  from  a  circular  disc  whose  radius 
is  /?;  required  the  centre  of  gravity  of  the  remaining  part,  when  the  hole  is 
tangent  to  the  circumference  of  the  disc. 

26.  Find  the  centre  of  gravity  of  the  frustum  of  a  pyramid  or  cone. 

It  will  be  in  the  line  which  joins  the  centre  of  gravity  of  the  upper  and 
lower  bases.  Let  h  be  the  length  of  this  line,  and  a  and  b  be  corresponding 
lines  in  the  lower  and  upper  bases  respectively,  required  the  distance,  measured 
on  the  line  A,  of  the  centre  from  the  lower  end. 

a^  +  2ab  +  36« 


Ans.  X  =  \h 


a"^  -rdb  +  b^ 


lib  —  0,  we  have  the  distance  of  the  centre  of  a  pyramid  or  cone  from  the 
base  equal  to  \h. 

27.  Find  the  centre  of  gravity  of  the  octant  of  a  sphere  in  which  the 
density  varies  directly  as  the  nth  power  of  the  distance  from  the  centre,  f 
being  the  radius  of  the  sphere. 

-      n  +  3        -      - 

28.  Find  the  centre  of  gravity  of  a  paraboloid  of  revolution  of  uniform 
density  whose  axis  ia  a.  __ 

Ans.  X  —  \a. 


108  GENERAL  PROPERTIES  [77-7d.J 

SOME   GENERAL   PROPERTIES   OF   THE   CENTRE   OF   GBi.VITY. 

77.  When  a  hody  is  at  rest  on  a  surface,  a  vertical  through 
the  centre  of  gravity  will  fall  within  the  support. 

For,  if  it  passes  without  the  support,  the  reaction  of  the  sur- 
face upward  and  of  the  weight  downward  form  a  statical 
couple,  and  rotation  will  result. 

78.  When  a  hody  is  suspended  at  a  point,  and  is  at  rest, 
the  centre  of  gravity  will  be  vertically  under  the  point  of  sus- 
pension. 

The  proof  is  similar  to  the  preceding.  When  the  preceding 
conditions  are  fulfilled  the  body  is  in  equilibrium. 

79.  A  body  is  in  a  condition  of  stable  equilibrium  when,  if 
its  position  be  slightly  disturbed,  it  tends  to  return  to  its  former 
position  when  the  disturbing  force  is  removed  ;  of  itnstoMe 
equilibrium  if  it  tends  to  depart  further  from  its  position  of 
rest  when  the  disturbing  force  is  removed ;  and  of  indifferent 
equilibrium  if  it  remains  at  rest  when  the  disturbing  force  is 
removed. 

Example. 

A  paraboloid  of  revolution  rests  on  a  hori- 
zontal plane ;  required  the  inclination  of  its 
axis. 

Let  P  be  the  point  of  contact  of  the  para- 
boloid and  plane,  then  will  the  vertical  through 
P  pass  through  the  centre  of  gravity  G,  and 
PG  will  be  a  normal  to  the  paraboloid. 

The  equation  of  a  vertical  section  through  the  centre  is  y^  = 
2^3?,  in  which  x  is  the  axis,  the  origin  being  at  the  vertex. 

Let    a  =  AX  =  the  altitude  of  the  paraboloid ; 
d  =  GPP  =  the  inclination  of  the  axis ; 
then,  AG  =  ^  a,  (see  example  28  on  the  preceding  page) ; 

AJ!f=%a-p; 
hence, 

tan  ^  -  ^  — ^  =  \/5  =  v/      -P 
which  will  be  positive  and  real  as  long  as  fa  exceeds^.    In 


[80-81.]  OF  THE  CENTRE  OF  GRAVITY.  109 

this  case  the  equilibrium  is  stable.  When  |a  exceeds  j?  it  will 
also  rest  on  the  apex,  but  the  equilibrium  for  this  position  is 
unstable.  When  ^a  =  p,  6  =  90°,  and  the  segment  will  rest 
only  on  the  apex.  When  ^a  is  less  than  j9,  tan  6  becomes 
imaginary,  and  hence,  this  analysis  fails  to  give  the  position  of 
rest;  but  by  independent  reasoning  we  find,  as  before,  that  it 
will  rest  on  the  apex,  and  that  the  equilibrium  will  be  stable. 

80,  I'^'  a  plane  material  section  the 
sum  of  the  products  found  by  midti- 
plying  each  elementary  mass  by  the 
sqybo/re  of  its  distance  from  an  axis, 
equals  the  sum  of  the  similar  products 
in  reference  to  a  parallel  axis  passing 
through  the  centre,  plus  the  mass  mul- 
tiplied by  the  square  of  the  distance 
between  the  axes.  ^'**-  ^ 

Let  AB  be  an  axis  through  the  centre, 
CD  a  parallel  axis, 
D  =  the  distance  between  AB  and  CDy 
dm  —  an  elementary  mass, 
yi  ==  the  ordinate  from  AB  to  m, 
y  =  the  ordinate  from  CD  to  m,  and 
M  =  the  mass  of  the  section. 
Then 

f={y,  +  Dy  =  y,^  +  22/,  i>  +  U^. 
Multiply  by  dm  and  integrate,  and  we  have 

ffdm  =fy?dm  +  ^Dfyydm  +  D^fdm. 
But  since  AB  passes  through  the  centre,  the  integral  of 
y^dm,  when  the  whole  section  is  included,  is  zero  (see  Eq.  715), 
and yjfm  =  M\  hence, 

fy'dm  =fyMm  +  MB^.  (83) 

Similarly,  if  dA  be  an  elementary  area,  and  A  the  total 
area,  we  have 

fi^dA  =fyi'dA  +  Ajy^. 

81.  In  any  plane  area,  the  sum  of  the  products  of  each  ele- 
mentary area  multiplied  by  the  square  of  its  distance  from  an 
axis,  is  least  when  the  axis  passes  through  the  centre. 


110  CENTRE  OF  MASS.  [82.] 

This  follows  directly  from  the  preceding  equation,  in  which 
the  first  member  is  a  minimum  for  D  =  0, 


CENTRE   OF   THE  MASS. 

82,  The  centre  of  the  mass  is  such  aj^oint  that,  if  the  whole 
mass  he  multiplied  hy  its  distance  from  an  axiSf  it  will  equal 
the  sum  of  the  products  found  hy  multiplying  each  elementary 
mass  hy  its  distance  from  the  same  axis. 
Let  m  =  an  elementary  mass; 
M  =  the  total  mass ; 
a?i>  yi>  and  Zi  be  the  respective  coordinates,  of  the  centre 

of  the  mass,  and 
x,y,  and  z  the  general  coordinates, 
then,  according  to  the  definition,  we  have 

Mxi  =  Xm^x ;  \ 

My,  =  XmyA  (84) 

Mzx  =  Xmz  ;  j 
which  being  the  same  as  (70)  shows  that  when  we  consider 
the  force  of  gravity  as  constant  for  all  the  particles  of  a  body, 
the  centre  of  the  mass  coincides  with  the  centre  of  gravity. 
This  is  practically  true  for  finite  bodies  on  the  surface  of  the 
earth,  although  the  centre  of  gravity  is  actually  nearer  the 
earth  than  the  centre  of  the  mass  is. 

If  the  origin  of  coordinates  be  at  the  centre  of  the  mass, 
we  have 

2mx  =  0 ;     i:my  =  0 ;     2mz  =  0 ;  i^^a) 

which  are  the  same  as  (71a). 


OHAPTEK  IV. 


NON-CONCURRENT   FORCES. 


83.  Equilibrium  of  a  solid  body  acted  upon  by  any 
number  of  forces  applied  at  different  points  and  acting 
in  different  directions. 


1 

^     X-z 

c 

/ 

B 

•r      / 

+  zV^ 

/ 

V-- 

/ 

A 

+x 

-X 

0 

T7" 

y           -x' 

y 

X 

*Y, 

^   X^-^ 

y/ 

y^z 

f 

-X 

A 

E 

■  •^    I 

F 


Fio.  64. 


Let  A  be  any  point  of  a  body,  at  which  a  force  F  is  applied, 
and  O  the  origin  of  coordinates,  which,  being  chosen  arbitrarily, 
may  be  within  or  without  the  body.  On  the  coordinate  axes 
construct  a  parallelopipedon  having  one  of  its  angles  at  0^  and 
the  diagonally  opposite  one  at  A, 

Let  the  typical  force  F\>q  in  the  first  angle  and  acting  away 
from  the  origin,  so  that  all  of  its  direction-cosines  will  be  posi- 
tive ;  then  will  the  sign  of  the  axial  component  of  any  force  be 
the  same  as  that  of  the  trigonometrical  cosine  of  the  angle  wliicb 
the  direction  of  the  force  makes  with  the  axis. 


112  GENERAL  EQUATIONS  [83.1 

Let  a  =  the  angle  between  i^and  the  axis  of  a?, 

Q  —.     a        a  a  a      a      a        n      a   y 

^  __.     (£         CI  u  a      a      a        a      ki   ^ 

then  will  the  X,  Y,  and  Z-components  of  the  force  i^be 
X=  7^  cos  a, 

r=i^cos;8, 

Z  =  i^cos  7. 

The  point  of  application  of  the  ^-component,  being  at  any 
point  in  its  line  of  action,  may  be  considered  as  at  D,  where  its 
action-line  meets  the  plane  y3.  At  jEJ  introduce  two  equal  and 
opposite  forces,  each  equal  and  parallel  to  Jl,  and  since  they 
will  equilibrate  each  other,  the  mechanical  effect  of  the  system 
will  be  the  same  as  before  they  were  introduced.  Combining 
the  force  -I-  X  at  i>  with  —  JT  at  A',  we  have  a  couple  whose 
arm  is  DJ^  =  y  =  the  y-ordinate  of  the  point  A.  This  couple, 
according  to  Article  54,  w^ill  be  negative,  hence,  its  moment  is 

-Xy. 

Hence,  a  force  +  JT  at  ^  produces  the  same  effect  upon  a 
body  as  the  couple  —  JTy,  and  a  force  +  JT  at  ^. 

At  the  origin  O  introduce  two  equal  and  opposite  forces, 
each  equal  to  X,  acting  along  the  axis  of  x.  This  will  not 
change  the  mechanical  effect  of  the  system.  Combining  —  J^ 
at  0  with  -f  X  at  jS",  w^e  have  the  couple  -r  ^2,  and  the  force 
+  X  remaining  at  0.  Hence,  a  single  force  -\-X  at  A  is 
equivalent  to  an  equal  parallel  force  at  the  origin  of  coordin- 
ates^ and  the  two  couples^ 

—  Xy  and  +  Xs. 
Treating  the  Y-component  in  a  similar  manner,  we  have  the 
force 

-\-  Y  at  the  origin^  and 
the  moments, 

4-  Yx  and  —  Yz  ; 
and  similarly  for  the  Z-component,  the  force 

+  Z  at  the  origin^  and 
the  moments, 

^  Zx  and  +  Zy. 
But  the  couples  4-  Zy  and  —  Yz,  have  the  common  axis  a, 


I84.J  OF   STATICS.  U3 

and  hence  are  equivalent  to  a  single  couple  which  is  equal  to 
the  ali^ebraic  sum  of  the  two  ;  and  similarly  for  the  othei-s ; 
hence,  the  six  couples  may  bo  reduced  to  the  three  following : 

Zy  —  J2,  having  x  for  an  axis 
Xz-Zx,       "       y    'i     "     " 
Yx-Xy,     "       3    "     "     " 

hence,  Jbr  the  single  force  F  acting  at  A  there  may  be  subszi- 
tuted  the  three  a/xial  components  of  the  force  acting  at  the 
origin  of  coordinates^  .and  three  pairs  of  couples  having  for 
their  axes  the  respective  coordinate  axes. 

If  there  be  a  system  of  forces,  in  which 
Fx^  F2,  JF^j  etc.,  are  the  forces, 
^ij  yij  ^ij  the  coordinates  of  the  point  of  application  of  Fi, 

^j  yaj  ^j  -^2> 

etc.,  etc.,  etc., 

Oi,  02,  Os,  etc.,  the  angles  made  by  i^,  i^J,  etc.,  respectively  witft 

the  axis  of  a?, 
A?  A>  A>  6tc.,  the  angles  made  by  the  forces  with  y,  and 
7i'  72>  73j  etc.,  the  corresponding  angles  with  z  ; 

then  resolving  each  of  the  forces  in  the  same  manner  as  above, 

we  have  the  axial  components 

X  =  Fi  cos  a^  +  F2  cos  a^-^r  F^  cos  Og  +  etc.  =  IIFcos  a ;  "j 
T^=  i^  cosA  +  ^  cos  A  +  ^  cos  /S3  +  etc.  =  SFcosiS',  I  (85) 
Z  =  i^  cos  7i  +  i^  cos  72  +  /s  cos  73  +  etc.  =  ^Fcos  7 ;  ; 
and  the  component  moments 

Zy  —  Y3  =  S(Fy  cos  7  -  i^  cos  /S)  =  Z  M 
Xs—  Zx  =  XiFz  0,0%  a  — Fx  cos  7)  =  Jf ;  V  {%(^) 

Yx^Xy=  S{Fx  cos  13  -  Fy  cos  a)  =  iT;  ) 
in  which  Z,  M,  and  iVare  used  for  brevity. 

RESULTANT  FORCE  AND  RESULTANT  COUPLE. 

84.  Let  i?  =  the  resultant  of  a  system  of  forces  concurring 
at  the  origin  of  coordinates,  and  having  the 
same  magnitudes  and  directions  as  those  of 
the  given  forces ; 
8 


114  DISCUSSION  OF  EQUATIONS.  [85.J 

a,  J,  and  c  =  the  angles  which  it  makes  with  the  axes  a?, 

y,  and  z  respectively  ; 
G  =  the  moment  of  the  resultant  couple  ; 
d,  e,  and  /=  the  angles  which  the  axis  of  the  resultant 
couple  makes  with  the  axes  x,  y,  and  s  re 
spectivelj ; 
then 

X  =  R  cos  a ;  \ 

r=^cos5;l  (87) 

Z  =  li  cos  G  ;  J 

L  =  G  cos  d ;  \ 

i¥  =  6^  cos  ^  ;  I  {SS) 

JSf^G  cos/   j 

If  a  force  and  a  couple,  equal  and  opposite  respectively  to 
the  resultant  force  and  resultant  couple,  be  introduced  into  the 
system,  there  will  be  equilibrium,  and  R  and  G  will  both  be 
zero.     Hence,  for  equilibrium,  we  have 

X=0;         r-=0;         Z  =  0;  (89) 

Z-0;        J/"=0;         i7"=0.  (90) 

85,  Discussion  of  equations  (87)  and  {SH). 

1.  Suppose  that  the  hody  is  perfectly  free  to  move  in  any 
manner. 

a.  If  the  forces  concur  and  are  in  equilibrium,  equations 
(87)  only  are  necessary,  and  are  the  same  as  equations  (60) ; 
hence,  we  will  have 

X=0,     r  =  0,     Z=:0. 
J.  If  i?  =  0  and   G  is  finite,  equations  {SS)  only  are 
necessary. 

c.  If  R  and.  G  are  both  finite,  then  all  of  equations  (87) 
and  {SS)  may  be  necessary. 
2  If  one  point  of  the  body  i^fixed^  there  can  be  no  trans- 
lation, and  equations  (88)  will  be  sufiicient. 

3.  If  an  axis  parallel  to  x  is  fixed  in  the  body,  there  may  bo 
translation  along  that  axis,  and  rotation  about  it ;  hence,  the 
1st  of  (87)  and  the  1st  of  (88)  are  suflScient. 


[80.1 


PROBLEMS. 


115 


4.  If  two  points  arefixed^  it  cannot  translate,  but  may  rotate ; 
and  by  taking  x  so  as  to  pass  through  the  two  points,  the  equa- 
tion Z  =  0  is  sufficient. 

5.  If  one  point  only  is  confined  to  the  plane  xy,  the  body 
will  have  every  degree  of  freedom  except  moving  parallel  to  2, 
and  hence,  all  of  equations  (87)  and  {S^)  are  necessary  except 
the  3d  of  (87). 

6.  If  three  points,  not  in  the  same  straight  line,  are  cpnfined 
to  the  plane  xy,  it  may  rotate  about  2,  but  cannot  move  parallel 
to  3;  hence,  the  1st  and  2d  of  (87)  and  the  3d  of  (88)  are 
necessary  and  sufficient. 

7.  If  two  axes  parallel  to  x  are  fixed,  the  body  can  move 
only  parallel  to  x,  and  the  1st  of  (87)  is  sufficient. 

8.  If  the  forces  are  parallel  to  the  axis  of  y,  thei*e  can  be 
translation  parallel  to  y  only,  and  rotation  about  x  and  z. 

0.  If  the  forces  are  in  the  plane  xy,  the  equations  for  equi- 
librium become 

X  =  ^F cosa  =  Ji  cos  a  =  0;\ 
T=  £Fco&^=  Bcosh  =  0 ;  I  (91) 

Yx  -  Xy  =  ^{Fxcosl3-FycoQa)=  0.  ) 

[Obs.  In  a  mechanical  sense,  whatever  holds  a  body  is  a  force.  Hence, 
when  we  say  "  a  point  is  fixed,"  or,  ''  an  axis  is  fixed,"  it  is  equivalent  to  in- 
troducing an  indefinitely  large  resisting  force.  Instead  of  finding  the  value  of 
the  resistance,  it  has,  in  the  preceding  discussion,  been  eliminated.  When  we 
say  "  the  body  cannot  translate,"  it  is  equivalent  to  saying  that  finite  active 
forces  cannot  overcome  an  infinite  resistance.] 


86,  Applications  of  equations  (91). 

a.  problems  in  which  the  tension  of 

I.  A  body  AJJ,  whose  weight  is  W, 
rests  at  its  lower  end  upon  a  perfectly 
smooth  horizontal  plane,  and  at  its 
upper  end  against  a  perfectly  smooth 
vertical  plane :  the  lower  end  is  pre- 
vented from  sliding  by  a  string  (JB. 
Determine  the  tension  on  the  string, 
and  the  pressure  upon  the  horizontal 
amd  vetiicaZ  planes. 


STRING  IS  INVOLVED.. 


116  EXAMPLES.  [86.3 

Take  the  origin  of  coordinates  at  t7,  the  axis  of  x  coinciding 
with  CB^  and y  with  AC,  x  being  positive  to  the  right,  and  y 
positive  upwards. 

Let  W  =  the  weight  of  the  body  whose  centre  of  gravity  is 
at  6^; 
B  =  the  reaction  of  the  vertical  wall,  and,  since  there 
is  no  friction,  its  direction  will  be  perpendicular 
to  ^6^; 
ir=  the  reaction  of  the  horizontal  plane,  which  will  be 
perpendicular  to  OB ; 
I  =  the  horizontal   distance  from   O  to  the  vertical 

through  the  centre  of  gravity; 
t  =  the  tension  of  the  string. 
The  forces  may  be  considered  positive,  and  the  sign  of  the 
component  of  the  force  will  be  that  of  the  trigonometrical 
function.  To  determine  the  angle  between  the  axis  of  +  a? 
and  the  force,  conceive  a  line  drawn  from  the  origin  of  co- 
ordinates parallel  to  and  in  the  direction  of  the  force,  then 
will  the  angle  be  that  swept  over  by  a  line  from  +  x  turning 
left-handed  to  the  line  thus  drawn.  The  origin  of  coordi- 
nates may  be  at  any  point,  and  the  origin  of  moments  at  any 
other  point. 

Taking  the  origin  of  coordinates,  and  the  origin  of  moments 
both  at  C^  we  have 

X=.^cosO°  -f-  ?5cosl80°  -f  IF  cos  270°  +  ircos90°  =0; 
I^=  ^  sin  0°  -}-  2^  sin  180°  -f  Fsin  270°  4-  iTsin  90^  =  0  ; 
Moments  =  -  R.AC  ^  t.O  -W.CJ)  -h  JSf.  CB  =  0. 
Reducing  gives 

R-t  =  0, 

-r+ir=o, 

-  E.AC-W.CD  +  mCB=0', 
or. 

B  =  t, 

TF=ir, 


EXAMPLES. 


117 


We  Bee  from  this  that  the  horizontal  plane  supports  the 
entire  weight  of  the  piece,  and  that  the  pressure  against  the 
wall  equals  the  tension  of  the  string. 

We  also  notice  that  the  forces  N  and  W  being  equal,  paral- 
lel and  opposite,  constitute  a  couple  whose  arm  is  DB ;  and 
this  must  be  in  equilibrium  with  the  couple  t^  CA,R\  the 
arm  being  CA,  hence  we  have  W.DB  =  t,  CA,  as  before. 

2.  A  ladder  rests  on  a  smooth  horizontal  plane  and  against  a 
vertical  wall,  the  lower  end  being  held  by  a  horizontal  string ;  a 
person  ascends  the  ladder,  required  the  pressure  against  the 
wall  for  any  position  on  the  ladder. 

3.  A  uniform  beam,  whose  length 
is  AB  and  weight  TT,  is  held  in  a 
horizontal  position  by  the  inclined 
string  CD,  and  carries  a  weight  P  at 
the  extremity ;  required  the  tension  of 
the  string. 


Fio.  56. 


^•Sg(^  +  w 


Ans.  t 

4.  A  prismatic  piece  AB  is  per- 
mitted to  turn  freely  about  the 
lower  end  A,  and  is  held  by  a 
string  CE\  given  the  position  of 
the  centre  of  gravity,  the  weight 

W  of  the  piece,  the  inclination  of 
the  piece  and  string,  and  the 
point  of  attachment  jE\  required 
the  tension  of  the  string,  and  the 
pressure  against  the  lower  end  of  the  beam  at  A. 

5.  A  heavy  apiece  AB  is 
supported  oy  two  cords 
which  pass  over  jpuUeys  G 
and  Jj,  and  have  weights 
Pi  and  P  attached  to  them; 
reqidred  the  inclination  to 
the  horizontal  of  the  line 
AB  joining  the  jpoints  of 
attachment  of  the  cord. 

(Consider  the  pulleys  as  reduced 
to  the  points  C  and  D. ) 


Fia.  57. 


118 


EXAMPLES. 


LS6.J 


Let  Gy  the  centre  of  gravity  of  AB^  be  on  the  line  join 
ing  the  points  of  attachment  A  and  B ; 
a  =  AG',  h  =  BG', 
i  =  the  angle  DCM\ 
3  =  the  inclination  of  BA  to  DG\ 
a  =  J)OA',B,iid^  =  GDB, 
Eesolving  horizontally  and  vertically,  we  have 
X=P,  cos  (180°  -  MCA)  +  P  cos  WDB  +  TTcos  270°=  0 ; 

=  -  Pi  cos  (a  -i)-\-P  cos  (/3  +  ^)  =  0  ;  {a) 

T=PiQm  {a  -  i)  +  Psin  (y8  +  i)  -  1F=  0.  {h) 

Taking  the  origin  of  moments  at  G,  we  have 

-  Pi  X  ^j?  +  P  X  ^i?i  +  Tr  X  0  =  0; 
or,  -  P^,a  sin  {a  +  h)  +  P  .  h  sin  (y8  -  S)  =  0.         (c) 

The  angle  i  is  given  by  the  conditions  of  the  problem ;  hence 
the  three  equations  {a\  (6),  and  (c)  are  sufficient  to  determine 
the  angles  a,  /3,  and  8,  when  the  numerical  values  of  the  given 
quantities  are  known.     The  inclination  will  be  8  +  i. 

6.  Suppose,  in  Fig.  58,  that  the  strings  are  fastened  at  6^ and 
P,  and  that  DO^  AC,  and  BJ)  are  given,  required  tlie  incli- 
nation of  AB. 

[The  solution  of  this  problem  involves  an  equation  of  the  8th  degree] . 

c  -.A         7.  A  heavy  piece  AB,  Fig.  59, 

is  free  to  swing  about  one  end  A, 
and  is  supported  by  a  string  BC 
which  passes  over  a  pulley  at  C,  and 
is  attached  to  a  weight  P ;  find 
the  angle  A  CB  when  they  are  in 
equilibrium. 

8.  A  weight  W  rests  on  a  plane 
whose  inclination  to  the  horizontal  is 
i,  and  is  held  by  a  string  whose  in- 
clination to  the  plane  is  6  ;  required 
the  relation  between  the  tension  P 
and  the  weight,  and  the  value  of  the 
normal  pressure  upon  the  planed 

Fig.  60 

Arts.  P  -= ^  TF;      JSiormal pressure  = ^     W. 


cos  6 


cos  6 


[86.] 


EXAMPLES. 


119 


h.    EQUILIBRIUM   OF    PERFECTLY    SMOOTH    BODIES    IN     CONTACt    WITH 

EACH    OTHER. 

9.  A  heavy  beam  rests  on  two  smooth  inclined  planes^  as  in 
Fig.  61 ;  required  the  inclination  of  the  heam  to  the  horizontal^ 
and  the  reactions  of  the  respective  planes. 

Let  AC  and  CB  be  the  in- 
cliued  planes;  AB  the  beam 
whose  centre  of  gravity  is  at  G, 
When  it  rests ^  the  reactions  of 
the  planes  must  be  normal  to  the 
planes,  for  otherwise  they  would 
have  a  component  parallel  to  the 
planes  which  would  produce  mo- 
tion. 

Let  a^  =  AG\    «2  =  (j^^ \ 
J2  =  the  reaction  at  A  ; 
B'=   "        "        "-g; 
W=  the  weight  of  the  beam  ; 
a  =  the  inclination  of  ^6^  to  the  horizon 


Fio.  01. 


Take  the  origin  of  coordinates  at  the  centre  of  gravity  O  of 
the  body,  x  horizontal  and  y  vertical. 
The  forces  resolved  horizontally  give 

X  =  i?  cos  (90°  -  a)  +  7?'  cos  (90°  -f  /3)  +  TFcos  270°  =  0  ;     {a) 
and  vertically, 

Y=zR  sin  (90°-  a)  -|-  R'  sin  (90°  +  /S)  +  If  sin  270°  =  0.     {h) 
The  moment  of  J?  sin   a   is,  .  .".  +  ^  sin  a  x  ^6^  sin  Q. 


"  R'  sin  ^  is,  . 
"  "  TFcos  90°  is,  . 
"         "  R  cos   a   is,  . 

"  R'  cos  /3  is, . 

Hence  Xy  —  Yx 


.  +  ^'  sin  y8  X  GB  sin  0, 
.       0. 

.  —  jS  cos  a  X  AG  cos  0. 
.  +  ^'  cos  /3  X  GB  cos  6, 
Ra^  sin  a  sin  ^  +  R'a^  sin  /9  sin  0 
r-  Ra^  cos  a  cos  Q  +  R'a^  cos  y9  cos  ^ 

=  —  Ra^  cos  (a  +  ^)  +  7?'fl^  cos  (/3-^=0.     (o) 


120  EXAMPLES.  [86.1 

It  is  generally  better  to  deduce  the  values  of  the  moments 
directly  from  tlie  definitions  ;  (see  Articles  51  to  67).  To  do 
this  in  the  present  case,  let  fall  from  G  the  perpendiculars  aO 
and  hG  upon  the  action-lines  of  the  respective  forces  ;  then 

hG  =  a^,  sin  (90°  _  (^  _  (9) )  --=  «2  cos  (^-6); 
aG  =  «i  sin  (90°  —  (a  +  ^)  )  =  «i  cos  {a  -\-  6); 

and  we  have 

the  moments  =  —  R* aG  +  B' »  hG  =  —  Ra^  cos  {a  -\-  &)  ->r 
R'a^  cos  (/3  —  ^)  =  0  ;  as  given  above. 

Solving  equations  (a),  (5),  and  (c),  we  find 

sm  (a  +  yS)      *  sm  (a  +  yS)      ' 

.       a       (^i  COS  a  sin  y8  —  «2  sin  a  cos  yS 

tan   U  =  7 r — : : tt . 

(%  +  a.^  sin  a  sin  p 

LfR  =  R\  then 

sin  /8  =  sin  a  ; 

which  are  the  conditions  necessary  to  make  the  normal  reactions 
equal  to  each  other. 

The  reactions  prolonged  will  meet  the  vertical  through  the 
centre  of  gravity  at  a  common  point  D,  and  if  the  beam  be 
suspended  at  D  by  means  of  the  two  cords  DA  and  DR  it 
will  retain  its  position  when  the  planes  AC  and  OR  are 
removed. 

If  y8  =  90°,  the  plane  OR  will  be  vertical,  and  we  find 

J^  =  TFseca;  ' R^  =  ^-^^W  =  Wtan  a; 

cos  a 

tan  0  = ■ cot  a. 

«!    +  «2 

li  ai  =  02,  then 

-        sin  (6  —  a)    ' 
tan  0  =  ^    ^         .  -i' 

2  sm  a.  sm  p 


[86.] 


EXAMPLES. 


121 


'/•-f 


Ays 


l{/3=  90°  and  a  =  0°,  then 

B'  =  0,  6  =  90°,  and  ^  =  TF. 

A  special  case  is  that  in  which  the  beam  coincides  with  one 
of  the  planes.     The  formulas  do  not  apply  to  this  case. 


10.  Two  equal,  smooth  cy- 
linders rest  on  two  smooth 
planes  whose  inclinations  are 
a  and  yS  respectively ;  required 
the  inclination,  6,  of  the  line 
joining  their  centres. 


Fio.  62. 

Ans,  Tan  6  =  ^(cot  a  —  cot  ff). 

11.  A  heavy,  uniform,  smooth  beam 
rests  on  one  edge  of  a  box  at  O,  and 
against  the  vei-tical  side  opposite ; 
required  its  inclination  to  the  vertical. 
Let  g  be  the  centre  of  gravity. 


Ans.  Sin  6 


=</ 


12.  Three  equal,  smooth  cylinders  are 
placed  in  a  box,  the  two  lower  ones  being 
tangent  to  the  sides  of  the  box  and  to  each 
other,  and  the  other  placed  above  them 
and  tangent  to  both ;  required  the  pressure 
aerainst  the  bottom  and  sides  of  the  box, 

^  Fio.  64. 

Ans.  Pressure  on  the  hottom  =  total  weight  of  the  cylin- 


Pressure  on  one  side  =  J  weight  of  one 
cylinder  x  tan  30°^ 

13.  Two  homogeneous,  smooth^prisma- 
tic  bars  rest  on  a  norizontal  plane,  and  are 
prevented  from  sliding  upon  it ;  required 
their  position  of  equilibrium  when  leaning 
against  each  other.  fio.  es. 

Let  AB  and  CD  be  the  two  bars,  resting  against  each  other 


aK 


EXAMPLES. 


at  B ;  then  will  they  be  in  equilibrium  when  the  resultant  of 
their  pressures  at  B  is  perpendicular  to  the  face  of  CD. 

Let  J   =  AB\  G  =  CD\  x^BD\ 

a   =  AD  =  the  distance  between  the  lower  ends  of  the 

bars ; 
W  =  the  weight  of  AB  ; 
W-^  =  the  weight  of  CD ; 

^aiid  (?  the  respective  centres  of  gravity  of  the  bars, 
which  will  be  at  the  middle  of  the  pieces ;  then  we  have 

2  {a^  ■\-h^-a?)  x'  W=  g  {a^  -V  +  a^  ("«'  +  ^'  +  ^)  ^^i ; 

which  is  an  equation  of  the  fifth  degree,  and  hence  always  ad- 
mits of  one  real  root. 

14.  The  upjper  end  of  a  heavy  piece 
rests  against  a  smooth^  vertical  plane^ 
and  the  lower  eyid  in  a  smooth^  spheri- 
cal howl ;  required  the  i^osition  of  equi- 
librium. 

Let  AB  be  the  piece,  BF  the  verti- 
cal surface,  EA  the  spherical  surface, 
and  g  the  centre  of  gravity  of  the  piece. 
When  it  is  in  equilibrium,  the  reaction  at  the  lower  end  will  be 
in  the  direction  of  a  normal  to  the  surface,  and  hence  will  pass 
through  C^  the  centre  of  the  sphere,  and  the  reaction  of  the  ver- 
tical plane  will  be  horizontal. 

Let  W  =  the  weight  of  the  piece  ; 

r  =  the  radius  of  the  sphere ; 

a=:  Ag;  b  =  Bg;  l  =  AB;  d=  CF\ 
B  —  the  reaction  of  the  vertical  plane ; 
N  =  the  reaction  of  the  spherical  surface ; 

i  =  the  inclination  of  the  beam  to  the  horizontal ; 

6  =  the  inclination  of  the  radius  to  the  horizontal. 

Take  the  origin  of  coordinates  at  ^,  x  horizontal  and  y  ver- 
tical ;  and  we  have 

X=ircos^  +  ^  cos  180°  +  Trcos270°  =  0; 
.  T=  iTsin  e  +  B  sin  180°  +  TTsin  270°  =  0 ; 


Fia.  66. 


Moments  =  4-  R.b  sin  i  —  iV^.a  sin  (^  —  i)  =  0 ; 


[86.]  INDETERMINATE  PROBLEMS.  123 

and  the  geometrical  relations  give, 

^  cos  *  =  KB  =  HF=  ^  +  /•  cos  0, 
From  these  equations,  we  have 

ir=  TTcosec  e\  R  =  TFcot  d\ 
a  sin  {6  —  i)  —  h  cos  6  sin  ^  =  0, 
which,  by  developing  and  reducing,  becomes 
{a  +  h)  tan  i  =  a  tan  6 ; 
this,  combined  with  the  fourth  equation  above,  will  determine 
^  and  6. 

The  position  is  independent  of  the  weight  of  the  piece,  but 
depends  upon  the  position  of  its  centre  of  gravity. 

15.  A  heavy  prismatic  bar  of  infinitesimal 
cross-section  rests  against  the  concave  arc  of  a  ^ 
vertical  parabola,  and  a   pin  placed   at   the  \;^' 
focus  ;  required  the  position  of  equilibrium.        \      X  \^     / 

Let  1=  AB  =  length  of  the  bar  ;  f^  CD      \^jfy 
—one-fourth  the  parameter  of  the  parabola,  O        ^     v 
being  tiie  focus,  and  6  =  A  CD.  w 

Fio.  67. 

16.  Eequired  the  form  of  the  curve  such  tliat  the  bar  will 
rest  in  all  positions. 

Ans.  The  polar  equation  is  r  =  il  -i-  o  sec  ^,  in  which  I  is 
the  length  or  the  bar,  and  c  an  arbitrary  constant.  It  is  the 
equation  of  the  conchoid  of  Nicomedes. 

C.    INDETERMINATE   PROBLEMS. 

17.  To  determine  the  jpresaurea  exerted  hy  a  door  upon  its 
hiiiges. 

Let  W  =  the  weight  of  the  door ; 

a  =  the  distance  between  the  hinges ; 
h  =  the  horizontal  distance  from  the  centre  of  gravity 
of  the  door  to  the  vertical  line  which  passes 
through  the  hinges ; 
F  =  the  vertical  reaction  of  the  upper  hinge ; 
Fi  =  the  vertical  reaction  of  the  lower  hinge ; 
H  =  the  horizontal  reaction  of  the  upper  hinge ; 
Bx  =  the  horizontal  reaction  of  the  lower  hin/;e ; 


124  INDETERMINATE  PROBLEMS.  [Sfi] 

then 

X=  H-  ZTi  =  0; 
Y=  F-\-  F^  -W  =  0; 
Xy  -  Yx  =  Ha  -Wl  =  0; 


which  give 


H=Hx  =  -W\  and 


F-{-  F^  =W. 

The  result,  therefore,  is  indeterminate,  but  we  can  draw  two 
general  inferences :  1st,  The  horizontal  pressures  upon  the 
hinges  are  equal  to  each  other  hut  in  opposite  directions  j  and, 
2d,  The  vertical  reaction  upon  both  hinges  equals  the  weight 
of  the  door. 

It  is  necessary  to  have  additional  data  in  order  to  determine 
the  actual  pressure  on  each  hinge.  The  ordinary  imperfec 
'tions  of  workmanship  will  cause  one  to  sustain  more  weight 
than  the  other,  but  as  they  wear  tliey  may  approach  an 
equality. 

The  horizontal  and  vertical  pressures  being  known,  the 
actual  pressures  may  be  found  by  the  triangle  of  forces.  If 
the  upper  end  sustains  the  whole  weight,  the  total  pressure 

pf 

upon   it   will   be  —  Va^  +  h^.     If  each   sustains   one-half  the 

weight,  the  pressure  on  each  will  be  one-half  this  amount. 

18.  A  rectangular  stool  rests  on  four  legs,  one  being  at  each 
corner  of  the  stool ;  required  the  pressure  on  each. 

(The  data  are  insufficient.)       • 

P 

19.  A  weight  P  is  supported  by  three  un- 
equally inclined  struts  in  one  plane  ;  required 
the  amount  which  each  will  sustain. 


[Obs.    If  more  conditions  are  given  than  there  are  quantities  to  be  deteiv 
mined,  they  will  either  be  redundant  or  conflicting.] 


[86.1 


EXAMPLES. 


125 


d*    STRESS   ON    FKAMF8 


20.  Sujpjpose  that  a  tricmgula^r  truss.  Fig.  69,  is  loaded  with 


/> 


0.    @    (^ 


equal  weights  at  the  upj>er  apices  ;  it  is  required  to  find  tJie 
stress  upon  any  of  the  pieces  of  the  truss, 

[The  stress  is  the  pull  or  push  on  u  piece.] 

Let  the  truss  be  supported  at  its  ends,  and  let 

I  =  Aa  =  ab  =  etc.,  =  the  equal  divisions  of  the  span  AJB ; 
iV=  the  number  of  bays  in  the  chord  AB ; 
L  —  Nl  —  AB^  the  span ; 

Pi,P2,Pa,  etc.,  be ihe  weights  on  the  successive  apices;  which 
we  will  suppose  are  equal  to  each  other ;  hence 
p  =pi  =p2  =  etc.; 
iVp  =  the  total  load  ; 
F=  the  reaction  at  ^;  and 


Fi  = 


B. 


1st.  TTiere  will  he  equilibrium  among  the  exterTial  forces. 

All  the  forces  being  vertical,  their  horizontal  components  will 
be  zero,  hence 

X=0; 

F=  Y+Y^-'Sp  =  V+Y^-Np  =  ^',  {a) 

and  taking  th^  origin  of  moments  at  B,  observing  that  the 
moment  of  the  load  is  the  total  load  multiplied  by  the  horizontal 
distance  of  its  centre  of  gravity  from  B,  we  have 

-V.AB  +  Np^iAB  =  0', 
or, 

V.L-Np,iL=0', 


126 


STRESS  ON 


[86.J 


which  in  {a)  gives  Y^  also  equal  to  ^Np ;  hence  the  6U]>i)ort9 
sustain  equal  amounts,  as  the^  should,  since  the  load  is  sy me- 
trical in  reference  to  them,  and  is  independent  of  the  forni  of 
trussing. 

2d.  To  determine  the  internal  forces. —  Conceive  that  the 
truss  is  cut  hy  a  vertical  plane  and  either  jpart  removed  while 
we  consider  the  remaining  part.  To  the  pieces  in  the  jplane 
section^  apjply  forces  acting  in  such  a  manner  as  to  jproduce  the 
same  strains  as  existed  before  they  were  severed.  Consider 
the  ^forces  thus  introduced  as  external^  and  the  problem  is 
reduced  to  that  of  determining  their  value  so  that  there  sJiaU 
he  equilibrium  among  the  new  system  of  external  forces. 


Let  CD,  Fig.  69,  be  a 
vertical  section,  and  suppose 
that  the  right-hand  part  is 
removed.  Introduce  the  ex- 
ternal forces  in  place  of  the 
strains,  as  shown  in  Fig.  70. 


Pio.  70. 


Let  H  =  the  compressive  strain  in  the  upper  chord ; 
ZTi  =  the  tensile  strain  in  the  lower  chord ; 
F  =  the  pull  in  the  inclined  piece  ; 
Q  =  the  inclination  of  i^to  the  vertical ; 
n  =  the  number  of  the   bay,  bD,  counting  from   A 

(which  in  the  figure  is  the  3d  bay) ;  and 
D  =  CD  =  the  depth  of  the  frame. 

^JThe  origin  of  coordinates  may  be  taken  at  any  point.    Take 
Rt  A,  X  being  horizontal  and  y  vertical. 

^     Eesolving  the  forces, 

Z=i?;cos0°+ircoal80°  +  Fcos  (270°  +  O4-rco8  90* +  2;?  cos  270'' =0; 
r^H.sin  0°-|-^sm  180°+2^sin  (370^+0°)+  Fsin  90°4-2p  sin  270°=0; 

or,  Bt-JI+FBin0=O;  (a) 

V-np  -Fcos0  =  O;  (b) 

and  the  moments, 

-  i7i'pl  +  ITD  -  F.  nlcoB  9  =  0.  (c) 


86.]  FRAMES.  127 

Eliminating  /'between  equations  {b)  and  (g\  substituting  the 
value  of  F  =  i^p,  and  reducing,  give 


ff=^n{]V-n);  {d) 


that  is,  the  strains  on  the  hays  of  the  ujpper  chord  vary  as  the 
product  of  the  segments  into  which  the  lower  chord  is  di/mded 
hy  th^e  joint  directly  under  the  hay  considered. 

From  {b)  we  have 

Fqoa  6  =  Y-njp  =  i(ir-  2n)^;  (e) 

and  since  6  is  constant,  the  stress  on  the  inclined  pieces  decreases 
uniformly  from  the  end  to  the  middle. 

At  the  middle  n  =  ^N',  and  i^=  0  ;  hence, /or  a  uniform 
load,  there  is  no  stress  on  the  central  hraces. 

If  F  were  considered  as  a  jpush,  equation  {e)  would  be 
'negative.  ^' 

Eliminating  ^and  i'^from  (a),  we  have 

E,=   ^N  {n-\)-n{n-V)\^  (/) 

For  forces  in  a  plane  the  conditions  of  statical  equilibrium 
give  only  three  independent  equations,  {a),  (h)  and  {c) ;  (or  Eqs. 
(91) );  hence,  if  a  plane  section  cuts  more  than  three  indepen- 
dent pieces  in  a  frame^  tJie  stresses  in  that  section  ai'e  indeter- 
minate, unless  a  relation  can  be  established  among  the  stresses, 
or  a  portion  of  them  be  determined  by  other  considerations. 

21.  If  ir=  7,i?  =i?i  =i?2  =  etc.  =  1,000  lbs.,  AB  =  56 
feet  and  D  =  4;  feet ;  required  the  stress  on  each  piece  of  the 
frame. 

22.  In  Fig.  69,  if  ^i  andj92  are  removed,  and  j^g  =  p^  =  p^  = 
1,000  lbs.,  find  the  stress  on  the  bay  2  —  3,  and  the  tie  2  —  h. 

23.  If  all  the  joints  of  the  lower  chord  are  equally  loaded, 
and  no  load  is  on  the  upper  chord,  required  the  stress  on  the  7i** 
pair  of  braces,  counting  from  A,  Fig.  69. 

An^,  ■i(iV  —  2n  +  1)  ^  sec  0, 


128 


STRESS   ON  A 


U 


24.  A  roof  truss  ADB  is  loaded  with  equal  weights  at  the 
equidistant  joints  1,  2,  3,  etc,  /  required  the  stress  on  any  of 
its  members, 

[Obs.  a  load  composed  of  equal  weights  on  all  the  joints  will  produce 
the  same  stress  as  that  of  a  load  uniformly  distributed,  except  that  the  latter 
would  produce  cross  strains  upon  the  rafters,  which  it  is  not  our  purpose  to 
discuss  in  this  place.] 

Let  the  tie  AB  be  divided  into  equal  parts,  Aa,  ab^  etc.,  and 
the  joints  connected  as  shown  in  the  figure.     The  joints  are 


assumed  to  be  perfectly  flexible.     The  right  half  of  Fig.  71 
may  be  trussed  in  any  manner  by  means  of  ties  or  braces,  or 
both,  and  yet  not  affect  the  analy- 
sis applied  to  the  left  half. 

Conceive  a  vertical  section  nm,  and 
the  right-hand  part  removed.  In- 
troduce the  forces  H^  Hi  and  F  as 
previously  explained,  and  the  condi- 
tions of  the  problem  will  be  repre- 
sented by  Fig.  72.  The  letters  of 
reference  given  below  involve  both  figures. 

Let  iV^=  the  number  of  equal  divisions  (bays)  in  ^^ ; 
n  =  the  number  of  the  bay  ho  counting  from  A ; 

I  =  Aa  =  ah,  etc. ; 
_p  =  the  weight  on  any  one  of  the  joints  of  the  rafter; 
V  =  the  vertical  reaction  at  ^  or  ^ ; 
D  ^=-  DC,  the  depth  at  the  vertex; 

e  =  J2c;  and*  =  DAC. 


Fro.  72. 


Then 

(iT. 


1)^  =  the  total  load;  .-.  F=  \{N -  \)jp. 


ROOF  TRUSS. 


129 


Take  the  origin  of  coordinates  at  -4,  and  the  omgin  of  mo- 
ments  at  the  joint  marked  2.  Resolving  the  forces  shown  in 
Fig.  72  horizontally  and  vertically,  we  have 

X=:  i^cos  (180°  +  aJ.l)-t-^i  cos  0°  +  i^"6in  (- 52c)  =  0  ; 
Y=  F-(/i-l)i?+JTsiu(180°  +  a^l)+i7i6inO°+i^co8(-  b2o) 
=  0; 
or,  —  Z^  cos  i  +  ^1  —  i^  sin  ^  =  0  ; 

V—  (7i-l)i>-^8in  i  +  FcoB  6  =  0; 

also  the  moments, 

HJ)2  -  V.Ab+  {n-l)p.i{n  -  2)  ^  =  0. 
But  from  Fig.  71  we  have 


h2_ 
CB 


AC 


{n-l)l 


Substituting  in  the  equation  of  moments  the  value  of  J2 
found  above,  of  F  =  i{J^  —  l)p,  of  Ab  =  {n  —  1)  Z,  and 
reducing,  give 


rr        ^^  /AT 


n  +  l)i>. 


By  means  of  the  other  two  equations,  and  {n  —  ij  tan  i  tan  6 
=1,  we  find  H  =  ^(ir  -  n)p  cosec  i ; 

F=  i{n  -l)^sec^. 


e.    8TBE88   m   A   LOADED   BEAM. 

25.  Suppose  that  a  beam  is 
firmly  fixed  in  a  wall  at  one 
end,  and  that  the  projecting 
end  is  loaded  with  a  weight 
P  /  required  the  forces  in  a 
vertical  section  mn^  Fig.  73. 

Take  the  origin  of  coordi- 
nates a,t  A,  X  horizontal  and  y 
vertical.  Take  the  plane  sec- 
tion perpendicular  to  the  axis 
of  X,    Without  assuming  to  know  the  directions  in  which  the 


Fia.  78. 


130  LOADED  COED.  [86.] 

forces  in  the  section  act,  we  may  conceive  them  to  be  resolved 
into  horizontal  and  vertical  components.  Let  i^be  the  typical 
horizontal  force,  then  will 

hence,  some  of  the  F-forces  will  be  positive,  and  the  others 
negative. 

Neglecting  the  weight  of  the  beam,  and  letting  T\  be  the 
sum  of  the  vertical  components  in  nni,  we  have 
Z=  Zi- P  =  0      .-.  ri  =  P; 

as  shown  in  the  figure. 

The  forces,  +  P  and  —  P,  constitute  a  couple  whose  arm  is 
Aa'y  and  since  the  F-forces ,  2iVQ  the  only  remaining  ones,  the 
resultant  of  the  +  F'^s  and  the  —  Fs  must  constitute  a  couple 
whose  moment  equals  P.Aa  with  a  contrary  sign. 

[Obs.  Investigations  in  regard  to  the  distribution  of  the  forces  over  the 
plane  section  belong  to  the  Resistance  of  Materials.] 


L^p^  I 


f,   LOADED    COED. 

26.  Suppose  that  a  perfectly  flexible,  inextensihle  cord  is 
fixed  at  two  points  and  loaded  continuously^  according  to  any 
law  /  it  is  required  to  find  the  equation  of  the  curve  and  the 
tension  of  the  cord. 

Assuming  that  equilibrium 
has  become  established,  we 
may  treat  the  problem  as  if  the 
cord  were  rigid,  by  consider- 
ing the  curve  which  it  assumes 
as  the  locus  of  the  point  of  ap-  ^o  ^ 

plication  of  the  resultant.     The 

resultant  at  any  point  will  be  in  the  direction  of  a  tangent  to 
the  curve  at  that  point ;  for  otherwise  it  would  have  a  normal 
component  which  would  tend  to  change  the  form  of  the  curve. 

Take  the  origin  of  coordinates  at  the  lowest  point  of-  the 
curve.     Let  a  be  any  point  whose  coordinates  are  x  and  y\ 

X.  =  the  sum  of  the  x-components  of  all  the  external  forces 
between  the  origin  and  a ; 

Y  =  the  sum  of  the  y-components ; 


u 


LOADED  CORD. 


131 


t  =  the  tension  of  the  cord  at  a ; 
to  =  the  tension  at  the  origin. 

Eesolving  the  tension  {£)  by  multiplying  it  by  the  direction-  ^ 
cosine,  we  ha\e 

dm 
t  -J-  =^  the  aj-component  of  ty  and 

dy        , 
t-f-  =  the  y-component 

For  the  part  Ca,  equations  (91)  become 

as         ' 


^y 


Tx  =  0. 


(») 


[Obs.  In  the  problems  whlcli  we  shall  consider,  the  third  of  these  equations 
will  be  mmecessary,  since  the  other  two  famish  all  the  conditions  necessary 
for  sohing  them.] 


Let  aU  the  applied  forces  he  vertical. 
Then  X  =  0,  and  the  first  two  of  equations  {a)  become 


4  +  t 


dx 


ds 
From  the  first  of  these  we  have 


(>) 


dub 
t  -r-  =  to  =  a  constant ; 


-V 


hence,  tlie  horizontal  component  of  the  tension  will  he  constant 
throughout  the  length  for  any  law  of  vertical  loading. 
From  the  second  of  (5),  we  have 


132 


PARABOLIC  CORD. 


[86.] 


hence,  the  vertical  corrijponent  of  the  tension  at  any  jpoint 
equals  the  total  load  between  the  lowest  point  and  the  jpoint 
considered. 


27.  Let  the  load  he  uniformly  dis- 
tributed over  the  horizontal. 

(This  is  approximately  the  condition  of  the 
ordinary  suspension  bridge. ) 

Let  w  =  the  load  per  unit  of  length, 
then 

1^=  —  wx\ 


and  (b)  becomes 


-tQ-\-  t 


dx 

ds 


^wx  +  t^  =  0. 
ds 

Eliminating  t  gives 

tody  =  wxdx'y 

and  integrating  gives 

Uy  =  lw3?  H- 

24 
w 


^ 


0); 


W 


id) 


hence,  the  curve  is  a  parabola  whose  axis  is  vertical,  and  whose 

2^ 
parameter  is  — ^.    The  parameter  will  be  constant  when  t^-—w 

is  constant ;  hence  the  tension  at  the  lowest  jpoint  will  be  the 
same  for  all  jparabolas  having  the  same  jparameter  and  the 
same  loadjper  unit  along  the  horizontal,  and  is  independent  oj 
the  length  of  the  curve. 

To  find  the  tension  at  the  lowest  point,  substitute  in  equa- 
tion {d)  the  value  of  the  coordinates  of  some  known  point. 
Let  the  coordinates  of  the  point  J.  be  ajj  and  y^ ;  then  (i)  gives 


_wx^ 


(^) 


[86.]  THE  CATENARY.  133 

To  find  the  tension  at  any  point  we  have  from  the  first  ol 
equations  (c)  and  the  Theory  of  Curves 

*-^»-^-*» — Si — -*"'  ^  +  ^ 


^y,^  ^  ^  da?  ^J> 

To  find  the  tension  at  the  highest  jpoint  A,  from  (d)  find 

■     dx       X,'  W 

substitute  in  (fi),  and  we  obtain 


tox. 


(To  find  to  by  the  Theory  of  Moments,  take  the  origin  at  A.  The  load  on 
a*!  will  be  wXi ,  and  its  arm  the  horizontal  distance  to  the  centre  of  gravity  of 
the  load,  or  ^i ;  hence,  its  moment  will  be  itoXi*.  The  moment  of  the  tension 
will  be  Uyi ;  hence, 

tot/i  =  iiDX*i  or  to  =  n — )  as  before.) 

The  slope  (or  inclination  of  the  curve  to  the  horizontal)  may 
be  found  from  equation  (g) ;  which  gives 

tan^  =  ?2^^ 

28.  The  Catenary.  A  catenary  is  the  curve  assumed  by  a 
perfectly  flexible  string  of  uniform  section  and  density,  when 
suspended  at  two  points  not  in  the  same  vertical.  Mechanically 
speaking  the  load  is  uniformly  distributed  over  the  arc,  and 
hence  varies  directly  as  the  arc. 

To  find  the  equation, 

let  w  =  the  weight  of  the  cord  per  unit  of  length ; 

.*.  Y=  —  ws  {s  being  the  length  of  the  arc) ; 
and  equations  {b)  become 


-t„+t^  =  0; 

-v,a  +  t^  =  0. 
ds 


(h) 


134  THE  OATENABY.  186.] 


Transposing  and  dividing  the  second  by  the  first,  gives 

and  differentiating,  substituting  the  value  of  ds  and  reducing, 
give  

"(2) 


'■  u 

d^ 
Integrating  gives 


TTTf 


or,  passing  to  exponentials,  gives 

/-_  ^    ,    kIVT^  _dy       ds^ 
'^  dx^  yf     ^  d^~  dx^  dx 

or,  1  + 


from  which  we  find 


dx 


(♦•) 


d^ ""  L^ "      dx\ ' 

.^»^-.    ^0    J; 

which  integrated  gives 

which  is  the  equation  of  the  Catenary. 
Eliminating  -j-  between  equations  (*)  and  (J),  we  find 


[8«.J  THE  CATENARY.  136 

the  intejrral  of  which  is 


(m) 


which  gives  the  length  of  the  curve. 

The  following  equations  may  also  be  found 


■w  ^}  t„  ^  y       i>„2  f ' 


f   f.'.':)e  - 


~  w  dx  • 

If  ^  =  the  inclination  of  the  curve  to  the  vertical,  then 
X  —  8  tan  Q  logg  cot  \Q. 

The  tensions,  ^  and  ^o?  ai'©  so  involved  that  they  can  be  de- 
termined only  by  a  series  of  approximations.  The  full  devel- 
opment of  these  equations  for  practical  purposes  belongs  to 
Applied  Mechanics. 

The  catenary  possesses  many  interesting  geometrical  and  mechanical  prop- 
erties, among  which  we  mention  the  following :  — 

The  centre  of  gravity  of  the  catenary  is  lower  than  for  any  other  curre  of 
the  same  length  joining  two  fixed  points. 

If  a  common  parabola  be  rolled  along  a  straight  line,  the  locus  of  the  focus 
will  be  a  catenary. 

According  to  Eq.  {k)  it  appears  that  if  the  origin  of  coordinates  be  taken 
directly  below  the  vertex  at  a  distance  equal  to  ^q  -^  ^>  the  constant  of  integra- 
tion will  be  zero.  (This  distance  equals  such  a  length  of  the  cord  forming  the 
catenary  as  that  Ite  weight  will  equal  the  tension  at  the  lowest  point  of  the 
curve).     A  horizontal  line  through  this  point  is  the  directrix  of  the  catenary. 

The  radius  of  curvature  at  any  point  of  the  catenary  equals  the  normal  at 
that  i>oint,  limited  by  the  directrix. 

The  tension  at  any  point  equals  the  weight  of  the  cord  forming  the  cate- 
nary whose  length  equals  the  ordinate  of  the  point  from  the  directrix. 

If  an  indefinite  number  of  strings  (without  i^eight)  be  suspended  from  a 
catenary  and  terminated  by  a  horizontal  liife,  and  the  catenary  be  then  drawn 
out  to  a  strsjight  line,  the  lower  ends  of  the  vertical  lines  will  be  in  the  arc  of 
a  parabola. 

If  the  weight  of  the  cord  varies  continuously  according  to 
any  known  law  the  curve  is  called  Catenarian, 


136 


LAW  OF  LOADING. 


186.] 


29.  To  determine  the  equation  of  the  Catenarian  curve  of 
uniform  deiuity  in  which  the  section  varies  directly  as  the 
tension,   ^  vx*^ 

Let  k  =  the  variable  section  ; 

8  =  the  weight  of  a  unit  of  volume  of  the  cord  ; 
c  =  ifye  ratio  of  the  section  to  the  tension ; 
then  (5^1^^^^^^!  ^     11,1/u.TvY-w 

which  substituted  in  (h)  and  reduced,  gives 


i^\L.^^(~i^^ 


hey 
for  the  required  equation. 


logg  sec  chx, 


g,   LAW   OF   LOADING. 

30.  It  is  required  to  find  the  law  of  loading  so  that  the 
action-line  of  the  resultant  of  the  forces  at  any  jpoint  shall  he 
tangent  to  a  given  curve. 

Assume  the  loading  to  be  of  uniform 
density,  and  the  variations  in  the  load- 
ing to  be  due  to  a  variable  depth.  In 
Fig.  76,  let  O  be  the  origin  of  coordi- 
nates ;  Z  =  aJ  =  the  depth  of  loading 
over  a  point  whose  abscissa  is  aj ;  6?  = 
the  depth  of  the  loading  over  the  ori- 
gin, and  8  =  the  weight  per  unit  of 
volume  of  the  loading,  then 
Y=z  -fhZdx', 


which  in  Eq.  (5)  gives 


-tc^  + 


t-  -0' 


Transposing,  and  dividing  the  latter  by  the  former,  gives 


dx       to^ 


[86.]  LAW  OF  LOADING  137 

which,  differentiated,  gives 

Bat,  from  the  Theory  of  Curves,  we  have 


('  -  a 


d^y       \         darj  sec*  i 

dx^  ~~  p  '~      p 

in  which  p  is  the  radius  of  curvature,  and  i  is  the  angle  between 
a  tangent  to  the  curve  and  the  axis  of  x.  From  these  we  read- 
ily find 

^  _U  sec'* 
I       p 

At  the  origin  p  z=:  p^^  i  =z  0,  and  Z  =  d;  which  values  sub- 
stituted in  the  preceding  equation  give 

^        ,    sec**  .  . 

/.  Z/  =  apQ .  (n) 

Discussion.  For  all  curves  which  have  a  vertical  tangent,  we 
have  at  those  points 

i  =  90° ;     .*.  sec  z  =  oo  ,  and,  if  p  is  finite 

Z=  oo; 

hence,  it  is  practically  impossible  to  load  such  a  curve  through- 
out its  entire  length  in  such  a  manner  that  the  resultant  shall 
be  in  the  direction  of  the  tangent  to  the  curve.  A  portion  of 
the  curve,  however,  may  be  made  to  fulfil  the  required  con- 
dition. 

Zet  the  given  cur79e  he  the  aro  of  a  circle  /  then  p  =  Poj  ^^^^ 
equation  (ti)  becomes 

Z  ^  d  sec'  *, 
from  which  the  upper  limit  of  the  loading  may  be  found.    For 


188  EXAMPLES  OF  A  [80.] 

small  angles  sec^^  will  not  greatly  exceed  unity,  and  hence,  the 
upper  limit  of  the  load  will  be  nearly  parallel  to  the  arc  of  the 
circle  for  a  short  distance  each  side  of  the  highest  point.  At 
the  extremities  of  the  semicircle,  i  =  90°,  and  Z  =^  cx> , 

If  the  given  curve  he  a  jpardbola,  we  find  Z  =  c?,  that  is,  the 
depth  of  loading  will  be  constant ;  or,  in  other  words,  uni- 
formly distributed  over  the  horizontal.  This  is  the  reverse  of 
Prob.'27. 

(The  principles  of  this  topic  may  be  used  in  the  construction  and  loading  of 
arches.) 

31.  Let  the  tension  of  the  cord  he  uniform. 

We  observe  in  this  case  that  the  loading  must  act  normally 
to  the  curve  at  every  point,  for  if  it  were  inclined  to  it,  the 
tangential  component  would  increase  or  decrease  the  tension. 

Let  J9  =  the  normal  pressure  per  unit  of  length  of  the  arc ; 
then  j9<^  =  the  pressure  on  an  element  of  length,  and  this  mul- 
tiplied by  the  direction-cosine  which  it  makes  with  the  axis  of 
a?,  and  the  expression  integrated,  give 

jpds  yj\  =    I  pdx  —  the  £C-component,  and 

/  pdi/  =  the  y-component  of  the  pressures, 
hence,  equations  (a),  p.  131,  become 

—  ^0  +   I  pdx  +  t  -^  =  0  ; 


Jpdy 


ds  ' 


differentiating  which,  give 

dx 


[86.]  nor:^iallt  pressed  arc.  139 

Transposing,  squaring,  adding  and  extracting  the  square 
root,  give 

that  is,  tJie  normal  jyressure  varies  inversely  as  the  radius  of 
curvature, 

1.  If  a  string  be  stretched  upon  a  perfectly  smooth  curved 
surface  by  pulling  upon  its  two  ends  the  normal  pressure  upon 
the  surface  will  vary  inversely  as  the  radius  of  curvature  of  the 
surface,  the  curvature  being  taken  in  the  plane  of  the  string  at 
that  point. 

2.  If  /9  be  constant  p  will  be  constant ;  hence,  if  a  circular 
cylinder  be  immersed  in  a  fluid,  its  axis  being  vertical,  the  nor- 
mal pressure  on  a  horizontal  arc  being  unifonn  throughout  its 
circumference,  the  compression  in  the  arc  will  also  be  constant. 

h.  THE  LAW  OF  LOADING  ON  A  NORMALLY  PRESSED  ARC  BEING 
GIVEN,     REQUIRED   THE   EQUATION   OF   THE   ARO       ^-J     ^  ' 

32.  T/ie  ties  of  a  suspension 
hridge  being  normal  to  the  curve 
of  the  cahle,  and  the  load  uni- 
form along  the  span^  required    £  ""f" 
the  equation  of  the  curve  of  fio.  tt. 
the  cable. 


B 


^-{■-(gVi)"-|Hi*V(4HJt 


the  origin  being  at  O,  x  horizontal  and  y  vertical 
dy 
dx 


If  tan  t  =  -T-,  and  p^  =  the  radius  of  curvature  at  the  vertex, 


then 

X  =  ipo  (1  +  cos*  ^)  sin  i, 

y  =  ipo  sin^  i  cos  i. 

(See  solution  by  Prof.  S.  W,  Robinson,  Journal  of  the 
FrankUn  Institute,  1863,  vol.  46,  p.  145  ;  and  its  application  to 
bridges  and  arches,  vol.  47,  p.  152  and  p.  361.) 


140  NORMALLY  PRESSED  ARC.  [86.] 

33.  A  perfectly  flexible^  inextensible  trough  of  indefinite 
length  is  filled  with  afiuid^  the  edges  of  the  trough  heing  par- 
allel and  supported  in  a  horizontal  plane  /  required  the  equa 
tion  of  a  cross  section. 

The  length  is  assumed  to  be  indefinitely  long,  so  as  to  elimin- 
ate the  effect  of  the  end  pieces.  The  pressure  of  a  fluid  against  a 
surface  is  always  normal  to  the  surface,  and  varies  directly  with 
the  depth  of  the  fluid.  The  actual  pressure  equals  the  weight 
of  a  prism  of  water  whose  base  equals  the  surface  pressed,  and 
whose  height  equals  the  depth  of  the  centre  of  gravity  of  the 
said  surface  below  the  surface  of  the  fluid.  The  problem  may 
therefore  be  stated  as  follows  : — Required  the  equation  of  the 
curve  assumed  hy  a  cord  fixed  at  two  points  in  the  same  hori- 
zontal^ and  pressed  normally  hy  forces  which  vary  as  the  verti- 
cal distance  of  the  point  of  application  helow  the  said  hori- 
zontal. 

Let  A  and  JS  be  the  fixed  points.  Take  the 
origin  of  the  coordinates  at  Z>,  midway  be- 
tween A  and  JB,  and  y  positive  downwards. 
Let  3  be  the  weight  of  a  unit  of  volume  ;  then 

p  z=  By,  which  in  equation  {o)  gives 
t  =  Byp,  and  for  the  lowest  point 
t  =  BDpQ ;  in   which  D  is  the  depth  of  the 
lowest  point  and  po  the  radius  of  curvature  at  that  point ; 

But  from  the  Theory  of  Curves  we  have 


H 


which  substituted  above,  and  both  sides  multiplied  by  dy,  maj 
be  put  under  the  form 


[86.]  HYDROSTATIC  TROUGH.  141 

the  integral  of  which  is 

But  -^  =  Ojfory  =  2>;   .-.  0=-~ ;  which  substituted 

and  the  equation  reduced  gives 

Squaring  and  reducing,  gives 

These  may  be  integrated  by  means  of  Ellvptio  Functions. 
Making  y  =  Z>  cos  <^,  and  c  =  j—,  they  may  be  reduced  to 
known  forms.     Using  Zegendre^s  notation,  we  have 

(See  Article  by  the  Author  in  the  Journal  of  the  Franklin 
Institute,  1864,  vol.  47,  p.  289.) 


%7J 


CHAPTER   V. 

RELATION    BETWEEN    THE  INTENSITIES    OF    FOKCES   ON    DIFFEEENT 
PLANES   WHICH   CUT  AN  ELEMENT. 

87,  Distributed  Forces  are  those  whose  points  of  applica- 
tion are  distributed  over  a  surface  or  throughout  a  mass.  The 
attraction  of  one  mass  for  another  is  an  example.of  the  latter, 
some  of  the  properties  of  which  have  been  discussed  in  the 
Chapter  on  Parallel  Forces ;  similarly,  when  one  part  of  a  body 
is  subjected  to  a  pull  or  push,  the  forces  are  transmitted 
through  the  body  to  some  other  part,  and  are  there  resisted  by 
other  forces.  If  the  body  be  intersected  by  a  plane,  the  forces 
which  pass  through  it  will  be  distributed  over  its  surface. 
Plaiies  having  different  inclinations  heing  passed  through  an 
element,  it  is  jprojposed  to  find  the  relation  between  the  inte7isi' 
ties  of  the  forces  on  the  different  planes. 

88.  Definitions.  Stresses  are  forces  distributed  over  a  sur- 
face. In  the  previous  chapters  we  have  assumed  that  forces 
are  applied  at  points,  but  in  practice  they  are  always  distrib- 
uted. 

A  strain  is  the  distortion  of  a  body  caused  by  a  stress. 
Stresses  tend  to  change  the  form  or  the  dimensions  of  a  body. 
Thus,  Si  pull  elongates,  2^  push  compresses,  a  twist  produces  tor- 
sion, etc.     (See  Resistance  of  Materials^ 

A  simple  stress  is  a  pull  or  thrust.  Stresses  may  be  com- 
pound, as  a  combination  of  a  twist  and  a  pull. 

A  direct  simple  stress  is  a  pull  or  thrust  which  is  normal  to 
the  plane  on  which  it  acts. 

A  pull  is  cemXdiQvedi  positive,  and  a  push,  negative. 

The  intensity  of  a  stress  is  the  force  on  a  unit  of  area,  if  it 
be  constant ;  but,  if  it  be  variable,  it  is  the  ratio  of  the  stress 
on  an  elementary  area  to  the  area. 

To  form  a  clear  conception  of  the  forces  to  wliich  an  element 
is  subjected,  conceive  it  to  be  removed  from  the  body  and  then 


[89,90.] 


EESOLVED  STRESSES. 


143 


Bubjected  to  such  forces  as  will  produce  the  same  strain  that  it 
had  while  in  the  body. 

89,  Formulas  foe  the  intensitt  of  a  stress.  Let  i^  be  a 
direct  simple  stress  acting  on  a  surface  whose  area  is  A,  and  j} 
the  measure  of  the  intensity,  then 


F 

wlien  the  stress  is  uniform,  and 


(92) 


jp  =  -T-j,  when  it  is  variable. 


If  the  stress  be  variable  we  will  assume  that  the  section  is  so 
small  that  the  stress  may  be  considered  uniform  over  its  sur- 
face. 

90.  Direct  stress  resolved.     Let   the   prismatic  element 
AB,  Fig.  79,  be  cut  by  an  oblique 
plane  BK     Let  the  stress  F  be  simple 
and  direct  on  the  surface  CB,  and 
^  =  the  normal  component  of  i^on 

DF; 
T  =  the  component  of  F  along  the 
plane  DE^  which  is  called 
the  tangential  comjponent  / 
0  =  FON  =  the  angle  between  the 
action-line  of  the  force  and 
a  normal  to  the  plane  DE^ 
and  is  called  the  obliquity 
of  the  plane  ; 
A  =  the  area  of  CB,  and  A'  that  of  DE, 

Then,  according  to  equations  (62),  we  have 

ir=  Fcose; 
T  =  i^'sin  e. 

From  the  figure  we  have 

A'  —  A  sec  ^, 
hence,  on  the  plane  DE^  we  have 


Fio.  791 


144-  SHEARING  STRESS.  [fll.j 


AT  7    •  V  ^  ^COS  e  2    O 

,     JSormal  intensity,  j[)n=-  —t,  =  -i >,  =  J>  cos^  0; 

^         jzL  see  (7 

Tangential  intensity,  jp^  =  -jj  = =p6m0cos6, 

-a.  ^  S6C  u 


(93) 


Pass  another  plane  perpendicular  to  J)£^,  having  an  obli- 
quity of  90°  —  0;  then,  accenting  the  letters,  we  have 
y„  =^sin2^; 


i?  <  =  J9  cos  ^  sm  ^.  )  ^    ^ 

This  result  is  the  same  as  if  a  direct  stress  acting  upon  a 
plane  perpendicular  to  (7j5,  having  an  obliquity  of  90°  --  ^  in 
reference  to  DJS,  be  resolved  normally  and  tangentially  to  the 
latter. 

Combining  equations  (93)  and  (94)  we  readily  lind 

that  is,  when  an  element  {or  hody)  under  a  direct  sinuple  stress 
is  intersected  hy  two  jplanes  the  sum  of  whose  obliquities  is  90 
degrees,  the  sum  of  the  intensities  of  the  noi^mal  comjponents 
of  the  stress  equals  the  intensity  of  the  direct  simple  stress,  and 
the  intensities  of  the  tangential  stresses  are  equal  to  each  other, 

91.  Shearing  stress.  The  tangential  stress  is  commonly 
called  a  shearing  stress.  It  tends  to  draw  a  body  side  wise 
along  its  plane  of  action,  or  along  another  plane  parallel  to  its 
plane  of  action.  Its  action  may  be  illustrated  as  follows : — 
Suppose  that  a  pile  composed  of  thin  sheets  or  horizontal  layers 
of  paper,  boards,  iron,  slate,  or  other  substance,  having  friction 
between  the  several  layers,  be  acted  upon  by  a  horizontal  force 
applied  at  the  top  of  the  pile,  tending  to  move  it  sidewise.  It 
will  tend  to  draw  each  layer  upon  the  one  immediately  beneath 
it,  and  the  total  force  exerted  between  each  layer  will  equal  the 
applied  force,  and  the  resistance  at  the  bottom  of  the  pile  will 
be  equal  and  opposite  to  that  of  the  applied  force.  If  other 
horizontal  forces  are  applied  at  different  points  along  the  ver- 
tical face  of  the  pile,  the  total  tangential  force  at  the  base  of 
the  pile  will  equal  the  algebraic  sum  of  all  the  applied  forces. 

A  shearing  stress  aiid  the  resisting  force  constitute  a  couple, 


[91.] 


SHEARING  STRESS. 


14£ 


and  as  a  single  couple  cannot  exist  alone,  so  a  pair  of  shearing 
stresses  necessitate  another  pair  for  equilibrium. 

When  the  direct  siraple  stresses  on  the  faces  of  a  rectangular 
jparallelojpijpedon  are  of  equal  intensity^  the  shearing  stresses 
will  he  of  equal  intensity. 

Let  Fig.  SO  represent  a  paralellopipedon  with  direct  and 
shearing  stresses  applied  to 
its  several  faces.  At  pre- 
sent suppose  that  all  the 
forces  are  parallel  to  the 
plane  of  one  of  the  faces, 
as  a^fe^  and  call  it  a  plane 
of  the  forces  ;  then  will  the 
planes  of  action^  which,  in 
this  case,  will  be  four  of  the 
faces  of  the  parallel opipe- 
don,  be  perpendicular  to  a 
plane  of  the  forces. 

If   the  direct  stress  +  F 
=  —  F^  and  +  F'  =■  —  F\  they  will  equilibrate  each  other. 
The  moment  of  the  tangential  force  jT,  will  be 

Pt  X  areafc  x  ah ; 
and  of  T' 

p't  X  area  ac  x  hf. 

The  couple  T.ab  tends  to  turn  the  element  to  the  right  and 
T'.hf  to  the  left,  hence,  for  equilibrium,  we  have 

jpt  X  areafc  x  ah  =  _p\  x  area  ac  x  hf; 
but  area  fcxah  =  area  ac  x  hf  =  the  volume  of  the  ele- 
ment, hence 


-Tr — 

^  ' 

F 

? 

..^ 

^=^- 

► 

f a 

iiji'ii         ^ 

■■ 

^      IP 

•/ 

^r- 

e 

if. — 

f 



/ 

■n 

-T 

>.    ^ 

-F 
Fzo.  £0. 


The  effect  of  a  jpair  of  shearing 
stresses  is  to  distort  the  element, 
changi7ig  a  rectangular  orde  into  a 
rhomhoid,  as  shown  in  Fig.  81. 

Direct  stresses  are  directly  opposed 

to  each  other  in  the  same  plane  or  on 

opposite  surfaces ;    shearing  stresses 

act  on  parallel  planes  not  coincident. 

10 


(96) 


Fia.  8L 


146 


NOTATION. 


[92, 93.1 


92,  Notation.  A  very  good  notation  was  devised  by  Co- 
riolis,  wliicli  lias  been  used  since  1837,  and  is  now  commonly 
employed  for  the  general  investigations  on  this  subject.  It  is 
as  follows : — 

Let  J9  be  a  typical  letter  to  denote  the  intensity  of  a  stress 
of  some  kind ;  p^  the  intensity  of  a  stress  on  a  plane  normal  to 
X  ;  j^axe  the  intensity  of  a  stress  on  a  plane  normal  to  x  and  in  a 
direction  parallel  to  x,  and  hence  indicates  the  intensity  of  a 
direct  simple  stress  /  and  p^  the  intensity  of  a  stress  on  a 
plane  normal  to  x  but  in  the  direction  of  y,  and  hence  indi- 
cates the  intensity  of  a  shearing  stress.  Or,  generally,  the  first 
sub-letter  indicates  a  n,ormal  to  the  plane  of  action  and  the 
second  one  the  direction  of  action.     Hence  we  have 


INTENSrrCES    OF   THE    FORCES 


parallel  to 


X 

y 

Z 

I>^ 

JPxy 

P^ 

Pyx 

JPyy 

JPyz 

-  on  a  plane  normal  to  ^ 

Pz^ 

JPzy 

y; 


s. 


If  direct  stresses  only  are  considered,  one  sub-letter  is  suffi- 
cient; as  j!?a.,j!?y,  orj?,. 

93,  Tangential  stress  resolved.  Let 
T  be  the  tano^ential  stress  on  the  riffht  sec- 
tion  AB  =  ^,  the  section  being  normal  to 
y,  then 

Let  CD  be  an  oblique  section,  normal  to 
the  axis  y' ;  x'  and  x  being  in  the  plane  of 
the  axes  y  and  y' ;  then  will  the  angle  be- 
tween y  and  y'  be  the  obliquity  of  the 
plane  CD.     This  we  will  denote  by  iyy'). 

Let  the  tangential  force  be  parallel  to  the  axis  of  x. 

Resolving  this  force,  we  have 

Normal  component  on  CD  =  T  sin  {^y') ; 
Tangential  component  on  CD  =  T  cos  (yy'). 


FiQ.  82. 


[W.] 


RESOLVED  STRESSES. 


147 


Dividing  each  of  these  by  area  CD  =  ^^  -j-  cos  (yy'),  we 
have 


Normal  intenMty  =  pytyr  = 


T  Bin  {yy')  cos 


Tangential  intensity  =  py/x' = 


area  AB 

T_coB\(jyy') 
area  AB 


-  =pyxan  iyy')  cos  (yy');^ 


=  Pyx  co8«  {yy'y, 


(97) 


and  for  a  tangential  stress  on   a  plane  normal  to  aj,  resolved 
upon  the  same  oblique  plane  CD,  we  have 


(98) 


If  the  tangential  stresses  on  both  planes  (one  normal  to  y,  and 
the  other  normal  to  x)  are  alike,  and  the  obliquity  of  the  plane 
CD  less  than  90°,  the  resultant  of  their  tangential  components 
will  be  the  difference  of  the  two  components,  as  given  by 
equations  (97)  and  (98) ;  that  is,  it  will  be  ^9^/^/  ""i^Va:'?  ^^^ 
the  normal  intensity  will  be  the  sum  of  the  components  as 
given  by  the  same  equations.  The  revei*se  will  be  true  in  re- 
gard to  the  dh^ect  stresses. 

C^-{ ,'  94.  ^^  ^  hody  he  subjected  to  a  direct  simjple  stress;  it  u 
required  to  find  the  stresses  on  any  two  planes  perpendicular 
to  one  another  and  to  the  plane  of  the  forces  ;  also  the  intensity 
of  the  stress  on  a  third  plane  perpendicular  to  the  plane  of  the 
forces ;  and  the  normal  and  tangential  components  on  that 
plane. 

Let  the  forces  be  parallel  to 
the  plane  of  the  paper ;  A  O  and 
OB,  planes  perpendicular  to  one 
another  and  to  the  plane  of  the 
paper,  having  any  obliquity  with 
the  forces.  Let  the  axis  of  x 
coincide  with  OB,  and  y  with 
A  O,  Let  AB  be  a  third  plane, 
also  perpendicular  to  the  plane 
of  the  paper,  cutting  the  other 
planes  at  any  angle.  Take  y'  perpendicular  to  AB  and  oc^ 
parallel  to  it  and  to  the  plane  of  the  paper. 


Fio.  88. 


148 


EESOLVED  STRESSES. 


P5.I 


Jl 


The  oblique  forces  may  be  resolved  normally  and  tangen- 

tially  to  the  planes  AO  and 
0£,  by  means  of  equations  (93) 
and  (94).  The  problem  will 
then  be  changed  to  that  shown 
in  Fig.  84,  in  which  one  set  of 
stresses  is  simple  and  direct,  and 
the  other  set  tangential ;  and, 
according  to  Article  91,  the  ^V 
intensity  of  the  shearing  stress 
on   the  two  planes   w^ill  be   the 


Pro.  84. 


same ;  hence,  for  this  case 

Pxy  —  Pyx- 

The  intensity  of  the  total  normal  stress  on  the  plane  AB 
will  be  the  sum  of  the  normal  components  given  by  equations 
(93),  (94),  (97)  and  (98),  and  the  total  tangential  stress  will  be 
the  sum  of  the  components  of  the  tangential  stress  given  by 
the  same  equations ;  hence 

P  v'y  =  Vx%  sin'  {yy')  +  pyy  cos'  {yy')  +  2pxy  sin  {yy)  cos  [yy') ;  )  .gg. 

Pyx  =-{Pxx~Pyy)-  sin  {yy)  cos  {yy)  -\-  p^  ■{  cos'  {yy')  -  sin'  {yy)  )■  -S^ 

The  resultant  stress  on  AB  will  be,  according  to  equation 
(46),  d  being  90°, 

Pr  =  '^¥y'7^fv7\  (100) 

and  the  inclination  of  the  resultant  stress  to  the  normal,  y\  will 
be 


tan  (^'y' 


(101) 


95.  Discussion  of  equations  (99). 

A.  Find  the  indination  of  the  j^lane  on  which  there  is  no 
tangential  stress. 

In  the  2d  of  equations  (99)  make^j^a--  =  0,  and  representing 
this  particular  angle  by  (yy'),  we  find 


tan  2(yy')  = 


2  sin  {yy')  cos  (yyQ       _       2/?^^ 


cos^  {yy')  -  sin^  {yy')        p^^  - p^ 


,  (102) 


195.1  PRINCIPAL  STRESSES.  149 

which  gives  two  angles  diflFering  from  each  other  by  90°,  or, 
the  planes  will  be  perpendicular  to  one  another. 

Hence,  in  every  case  of  a  direct  simple  stress  upon  a  pair 
of  planes  perpendicular  to  one  another  and  to  a  plane  of  the 
stresses^  there  are  two  planes,  also  perpendicular  to  one  another 
and  to  thAi  plane  of  the  stresses,  on  which  there  is  no  tangential 
stress. 

These  two  directions  are  Q,2)\QiA  principal  axes  of  stress. 

Principal  a^oces  of  stress  are  the  normals  to  two  planes  per- 
pendicular to  one  another  on  which  there  is  no  tangential  stress. 

Principal  stresses  are  such  as  are  parallel  to  the  principal 
axes  of  stress.  (In  some  cases  there  is  a  third  principal  stress 
perpendicular  to  the  plane  of  the  other  two.) 

The  formulas  for  the  stresses  become  most  simple  by  refer- 
ring them  directly  to  the  principal  axes. 

a.  Let  one  of  the  direct  stresses  he  zero. 
Equation  102  gives 

tan  2(yyO  =  ^  (103) 

h.  Let  one  of  the  direct  stresses  he  a  pull,  and  the  other  a 
piLsh. 

Then 

tan  2(yy')  =      ^-^^  (104) 

Pxx   +  Pw 

c.  Let  them  act  in  opposite  senses  and  equal  to  each  other. 
Then 

tan2(YY')=^.  (105) 

d.  Let  there  he  no  tangential  stress  on  the  original  pianos^ 
orp^  =  0.        _ 

Then, 

tan  2(yyO  =  ^ ;     •*•  (yy')  =  0  or  90° ; 
and  the  original  planes  ^xq  principal  planes. 


160  RESOLVED  STRESSES.  [95.] 

e.  Let  there  he  no  direct  stresses. 
Then, 

tan  2(yy')  =  oo  ;  or  (yyO  =  45°  or  135° ;       (106) 

that  is,  if  on  two  planes,  perpendicular  to  one  another  and  to  the 
plane  of  the  stresses,  there  are  no  direct  stresses,  then  will  the 
stress  on  two  jplanes,  perpendicular  to  one  another  and  to  the 
plane  of  the  stresses,  whose  inclination  with  the  original  planes 
is  45°,  he  simple  and  direct. 

f.  Let  the  direct  stresses  he  equal  to  one  another  and  act  in 
the  same  sense,  and  let  there  he  no  shearing  on  the  original 
planes. 

Then 

tan2(YY')=-2; 

and  (yy')  is  indeterminate;  hence,  in  this  case  every  plane 
perpendicular  to  a  plane  of  the  stress  will  be  a  principal  plane. 

Examples. 

1.  A  rough  cube,  whose  weight ''is  550  pounds,  rests  on  a 
horizontal  plane.  A  stress  of  150  pounds  applied  at  the  upper 
face  pulls  vertically  upward,  and  another  direct  stress  of  125 
pounds,  applied  at  one  of  the  lateral  faces,  tends  to  draw  it  to 
the  right,  while  another  direct  stress  of  50  pounds  tends  to 
draw  it  to  the  left ;  required  the  position  of  the  planes  on  which 
there  are  no  tangential  stresses. 

If  the  cube  is  of  finite  size  it  will  be  necessary  to  modify  the 
problem,  in  order  to  make  it  agree  with  the  hypothesis  under 
which  the  formulas  have  been  established.  The  force  of  gravity 
being  distributed  throughout  the  mass,  would  cause  a  variable 
stress,  and  the  surface  of  no  shear  would  be  curved  instead  of 
plane.  We  will  therefore  assume  that  the  cube  is  without  weight, 
and  the  550  pounds  is  applied  directly  to  the  lower  surface. 
Then  the  vertical  stress  will  be  150  pounds,  the  remaining  400 
pounds  being  resisted  directly  by  the  plane  on  which  it  rests, 
and  so  far  as  the  present  problem  is  concerned,  only  produces 
friction  for  resisting  the  shearing  stress.     The  direct  horizontal 


[95.1  EXAMPLES.  //-//--  ^^^ 

Stress  will  be  50  pounds,  the  remaining  75  pounds  producing  a 
shearing  on  the  horizontal  plane.  The ^Qtm^v  force  tends  to 
turn  the  cube  right-handed  by  rotating  it  about  the  lower  right- 
hand  corner,  thus  producing  a  reaction  or  vertical  tangential 
stress  of  75  pounds.  Let  the  area  of  each  face  of  the  cube  be 
unity,  then  we  have  ^ 

Pxy  =  ^^  pounds ;    p^^  =  50  pounds  ;    j?j^  =s  150  pounds ; 

and  these  in  (102)  give 

.-.  (yt')  =  +  28°  9'  18'',  or  -  61°  50'  42". 

If  the  body  be  divided  along  either  of  these  planes,  the 
forces  will  tend  to  lift  one  part  directly  from  the  other  without 
producing  sliding  upon  the  plane  of  division. 

2.  A  rough  body,  whose  weight  is  100  pounds,  rests  on  an 
inclined  plane  ;  required  the  normal  and  tangential  components 
on  the  plane.     (Use  Eq.  (93).) 

3.  A  block  without  weight  is  secured  to  a  horizontal  plane 
and  thrust  downward  by  a  stress  whose  intensity  is  150  pounds, 
and  pulled  towards  the  right  by  a  stress  whose  intensity  is  150 
pounds,  and  to  the  left  with  an  intensity  of  100  pounds ;  re- 
quired the  plane  of  no  shear. 

4.  A  cube  rests  on  a  horizontal  plane,  and  one  of  its  vertical 
faces  is  forced  against  a  vertical  plane  by  a  stress  of  200  pounds 
applied  at  the  opposite  face,  and  on  one  of  the  other  vertical 
faces  is  a  direct  pulling  stress  of  75  pounds,  which  is  directly 
opposed  by  a  stress  of  50  pounds  on  the  opposite  vertical  face  ; 
required  the  position  of  the  plane  of  no  shear. 

In  this  case  the  weight  of  the  cube  would  be  a  third  princi- 
pal stress,  but  it  is  eliminated  by  the  conditions  of  the  problem. 
The  shearing  stress  is  25  pounds;  and  because  the  direct 
stresses  are  unlike,  we  use  Eq.  (104). 

5.  A  rectangular  parallelopipedon  stands  on  a  horizontal 
plane,  and  on  the  opposite  pairs  of  vertical  faces  tangential 


152  PLANES  OF  pW.i 

stresses  of  equal  intensities  are  applied ;  required  the  position 
of  the  plane  of  no  shear.     (See  Eq.  (106).) 

6.  In  the  preceding  problem  find  the  intensity  of  the  direct 
stress  on  the  plane  of  no  shear.  (Substitute  the  proper  quanti- 
ties in  the  1st  of  (99).) 

B.  To  find  the  planes  of  action  for  maxiniuin  and  minimum 
normal  stresses,  and  the  values  of  the  stresses. 

Equate  to  zero  the  first  differential  coefficient  of  tlie  1st  of 
Equations  (99),  and  we  have 


2i>aa:  COS  {yy')  sin  {yy')  -  2^^^  sin  (yy')  cos  (yy') 

[yy') 


—  2j?ay  sin^  iyy')  +  ^p^  cos^  (yy')  =  0 
.-.  tan  2(yy') 


I      (107) 


Pxx       JPyy 


which,  being  the  same  as  (102),  shows  that  on  those  planes 
which  have  no  shearing  stress,  the  direct  stress  will  be  either  a 
maximum  or  a  minimum.  Testing  this  value  by  the  second 
differential  coefficient,  we  find  that  one  of  the  values  of  (yy') 
gives  a  maximum  and  the  other  a  minimum. 

Comparing  (107)  with  the  second  of  (99),  shows  that  the  first 
differential  coefficient  of  the  value  of  the  direct  stress  on  any 
plane  equals  the  shearing  stress  on  that  jplane. 


From  (107),  observing  that  cos  (yy')  —  -/l  —  sin^  (yy')^  we 
find 


(108) 


and  these  values  in  the  1st  of  (99),  and  the  maximum  and 
minimum  values  designated  by  ^y/,  give 

i>Y'  =  \{V^-^J[>vv)  ±   ^/\\{P^  -Pyyf  +ya^h         (109) 

in  which  the  upper  sign  gives  the  maximum,  and  the  lower  the 
minimum  stress.  These  are  jprincip>al  stresses,  and  we  denote 
them  by  one  sub-letter. 


[«5.]  MAXIMUM  STRESS.  163 

a   Ifpxy  —  ^>  we  have 

(^y')  z=  0°  or  90°,  as  we  should. 


maximum,  p^,  =  i^^^  +  ^/ii^^w:  +  p^xv  \ 
minimum, p^  =  i^«,  -  \/^p^^+jp^^:, 


;} 


(110) 


hence,  the  maximum  normal  stress  will  be  of  the  same  hind  as 
the  principal  direct  stress, ^aa;  that  is,  if  the  latter  is  Q,pull, 
the  former  will  also  be  a  j>uU,  and  the  minimum  principal 
stress  will  be  of  the  opposite  kind. 

c.  If  there  are  no  direct  stresses  j[>^  will  also  he  zero,  and  we 
have 

(YY')  =  45°orl35°; 
and  * 

maximum  jp^i  =-Pxu  =  ~!Px'  for  minimum  / 

that  is,  the  principal  stresses  will  have  the  same  intensity  as 
the  shearing  stresses,  and  act  on  planes  perpendicular  to  one 
another,  and  inclined  45°  to  the  original  planes. 

Examples. 

1.  Suppose  that  a  rectangular  box  rests  on  one  end,  and  that 
one  pair  of  opposite  vertical  sides  press  upon  the  contents  of 
the  box  with  an  intensity  of  20  pounds,  and  the  other  pair 
of  vertical  faces  press  with  an  intensity  of  40  pounds,  and  that 
horizontal  tangential  stresses,  whose  intensities  are  10  pounds, 
are  applied  to  the  vertical  faces,  one  pair  tending  to  turn  it  to 
the  right,  and  the  other  to  the  left ;  required  the  position  of 
the  vertical  planes  of  no  shearing,  and  the  maximum  and 
minimum  values  of  the  direct  stresses. 

2.  For  an  application  of  Equations  (103)  and  (110)  to  the 
stresses  in  a  beam,  see  the  Author's  Resistance  of  Materials, 
2d  edition,  pp.  236-240. 

C.  To  find  the  position  of  the  jplan^s  of  maximum  and 
minimum  shearing. 


154  PRINCIPAL  STRESSES.  195..I 

Equate  to  zero  the  first  differential  coefficient  of  the  second 
of  (99)  and  reduce,  denoting  the  angles  sought  by  {YY'),  and 
we  find, 

-  cot2(Zr')  =  tan2(YY'); 

.-.  2rr'  =  2(YY')  +  90°; 

or, 

YY'  =  yy'  +  45° ; 

that  is,  the  planes  of  maximum  and  minimum  shear  mxike 
angles  of  4:6  degrees  with  the  pkincipal  planes. 

D.  Let  the  planes  he  principal  sections. 

Then  the  stresses  will  be  principal  stresses,  and  p^gy  =  0. 
Using  a  single  subscript  for  the  direct  stresses,  equations  (99) 
become 

py'  =  p^  sin^  {yy')  +  p^  cos^  (yy') ;)  ^ 

fyx   =  (Px  -  Pv)  sin  (yy')  cos  {i/y'). ) 

a.  Tjetp^  =  Vvj  ^^^^^ 

I>p'  =J>x  ;  and  p^^  =  0; 

that  is,  when  two  principal  stresses  are  alike  and  equal  on  a 
pair  of  planes  perpendicular  to  the  plane  of  the  stresses,  the 
normal  intensity  on  every  plane  perpendicular  to  the  plane  of 
the  stresses  will  he  equal  to  that  on  the  principal  planes,  and 
there  will  he  no  shearing  on  any  plane. 

This  condition  is  realized  in  a  perfect  fluid,  and  hence  very 
nearly  so  in  gases  and  liquids,  since  they  offer  only  a  very  slight 
resistance  to  a  tangential  stress.  If  a  vessel  of  any  liquid  be 
intersected  by  two  vertical  planes  perpendicular  to  one  another, 
the  pressure  per  square  inch  will  be  the  same  on  both,  and  will 
be  normal  to  the  planes ;  hence,  according  to  the  above,  it  will 
be  the  same  upon  all  planes  traversing  the  same  point.  This 
is  only  another  way  of  stating  the  fact  that  fluids  press  equally 
in  all  directions. 

h.  To  find  the  planes  on  which  there  will  he  no  normal pres- 
mire. 

For  this_^y/  in  (111)  will  be  zero ; 


[98.]  PROBLEM.  155 

.-.  tan  (yr/)  =  y/^  ^/^ZT; 

which,  being  imaginary,  shows  that  it  is  impossible  when  the 
stresses  are  alike  ;  but  if  thej  are  unlike,  we  have 

Jfj>x=  —jpm  t^6^  (yy')  =  ^^°y  and  the  2d  of  (111)  gives 

which  shows  that  when  the  direct  stresses  are  unlike  and  of 
equal  intejisity  on  planes  perpendicular  to  one  another,  the 
shearing  stress  on  a  plane  cutting  both  the  others  at  an  angle 
of  45  degrees,  will  be  of  the  same  intensity. 

Let  (yy')  =  45°,  or  135°,  then  (111)  become 

— ^-   .  JPi/x'  =  ^  iiPx  -2^v) ; ) 

in  the  latter  of  which  the  upper  sign  gives  a  maximum,  and  the 
lower  a  minimum  value. 

Using  the  upper  sign,  we  find 


\  \  (113) 


96.  Problem.  Find  the  plane  on  which  the  obliquity  of  the 
stress  is  greatest,  the  intensity  of  that  stress,  and  the  angle  of 
its  obliquity. 

Let  the  stresses  be  jprinaipal  stresses  and  of  the  same  hind, 
and  <^  the  angle  of  obliquity  of  the  required  plane  to  the  stress ; 
then 

8in<^  ^^'^  ,  th^  intensity  =  V{p»JPv)\  ^^^  t^®  angle  be- 

tween  the  principal  plane  x  and  the  required  plane  =  45°—  J^. 


156  CONJUGATE  STRESSES.  [97.1 

If  the  principal  stresses  are  unlike^  then 

sin  <^'  =^^ — i-^\  the  intensity  =  ^  —pxPvj  ^^^  *^®  angle  be- 
tween  the  principal  plane  a?,  and  the  oblique  plane  =  45°  —  ^<^'. 

Example. 

If  a  body  of  sand  is  retained  by  a  vertical  wall  and  the 
intensity  of  the  horizontal  push  is  25  pounds,  and  of  the  ver- 
tical pressure  is  75  pounds ;  required  the  plane  on  which  the 
resultant  has  the  greatest  obliquity,  and  the  intensity  of  the 
stress  on  that  plane. 

CONJUGATE    STRESSES. 

97.  A  pair  of  stresses,  each  acting  parallel  to  the  plane  of 
action  of  the  other,  and  whose  action-lines  are  parallel  to  a 
plane  which  is  perpendicular  to  the  line  of  intersection  of  the 
planes  of  action,  are  called  conjugate  stresses. 

Thus,,  in  Fig.  85,  one  set  of  stresses 
acts  on  the  plane  YY,  parallel  to  the 
plane  XX,  and  the  other  set  on  XX, 
parallel  to  YY.  In  a  rigid  body  the 
intensities  of  these  sets  of  stresses  are 
independent  of  each  other;  for  each 
set  equilibrates  itself.  Princijpal 
stresses  are  also  con  jug  ate. 

There  may  be  three  corijugate  stresses 
in  a  body,  and  only  three.  For,  in 
Fig.  85,  there  may  be  a  third  stress  on  the  plane  of  the  paper, 
which  may  be  parallel  to  the  line  of  intersection  of  the  planes 
XX  and  YY^  and  each  stress  will  be  parallel  to  the  plane  of  the 
other  two.  A  fourth  stress  cannot  be  introduced  which  will  be 
conjugate  to  the  other  three. 

Conjugate  stresses  may  be  resolved  into  normal  and  tangen- 
tial components  on  their  planes  of  action,  and  treated  according 
to  the  preceding  articles.  The  fact  that  the  stresses  have  the 
same  obliquity,  being  the  complement  of  the  angle  made  by 
the  planes,  simplifies  some  of  the  more  general  problems  of 
stresses. 


[98.] 


GENERAL  PROBLEM. 


157 


GENERAL   PROBLEM. 

98.  Given  the  st^^esses  on  the  three  rectangular  coordinate 
planes  /  required  the  stresses  on  any  ohligue  plane  in  any  re- 
quired direction. 

As  before,  the  element  is  supposed  to  be  indefinitely  small. 
Let  ahc  be  the  oblique  plane,  the 
normal  to  which  designate  by  n. 
The  projection  of  a  unit  of  area  of 
this  plane  on  each  of  the  coordinate 
plains,  gives  respectively 

cos  {nx)^  cos  (;iy),  cos  ijiz). 
■   The  direct  stress  parallel  to  x 
acting  on  the   area  cos  {naS)  will     „ 
give  a  stress  of  jt?aa;  cos  {nx)^  and        '  fio.  86. 

the  tangential  stress  normal  to  y  and  parallel  to  x  will  produce 
a  stress ^y:B  cos  {ny\  and  similarly  the  tangential  stress  normal 
to  2  and  parallel  to  x  gives ^^a.  cos  {nz) ;  hence  the  total  stress 
on  the  unit  normal  to  n  and  parallel  to  x  will  be 

i?na  =  Jpxx  COS  (tW?)  +  fy^  COS  {mj)  +  Jp^  COS  (n^) ; ' 


(114) 


similarly, 

Pnv  =  Pxv  COS  {rvx)  +jpyy  COS  {ny)  +  p^,  cos  {m)  ; 
p^  =  2)x»  cos  {nx)  +py»  cos  {ny)  +  p„  cos  {nz). 

Let  these  be  resolved  in  any  arbitrary  direction  parallel  to  s. 
To  do  this  multiply  the  first  of  the  preceding  equations  by  cos 
(«c),  the  second  by  cos  {sy),  and  the  third  by  cos  {sz),  and  add 
the  results. 

For  the  purpose  of  abridging  the  formulas,  let  cos  {nx)  be 
written  Qnx,  and  similarly  for  the  others.     Then  we  have 
Pnt  —  pxxCnxCsx  +  Pi/yCnyCsy  +  ^«  CnzCsz  \ 

+Pyz  {OnyQsz  +  CnzCsy)  -^pJ^nzCsx  I  (115) 
4-  GtixCsz)  +  pxy  (CnxCsy  +  CnyCsx).  ) 
This  expression  being  typical,  we  substitute  x  for  n  and  s, 
and  thus  obtain  an  expression  for  the  intensity  on  a  surface 
normal  to  x'  and  parallel  to  x'.  Or  generally,  substitute  suc- 
cessively x',  y\  z'  for  n  and  «,  and  we  obtain  the  following 
formulas ; 


168  GENERAL  FORMULAS.  [M.] 


DIRECT   STRESSES. 

+  ^pgJu2x'Gxx'-\-  '^^xyOxx'Cyx  ; 

i>yv  =I>xx0^xy'+  pyy(yyy' + Pzz^W  +  2^y,Cyy'Csy' 
+  'ijPzxOzy'  Oxy'-\-  ^j>^Cxy'Qyy' ; 

+  ^^pgxOzz'Qxz'  +  ^jp^Qxz'Oyz' ; 

TANGENTIAL   STRESSES. 

jpy/e'  =_^««Ca3y'Ca?3'  +  :PvyCy^JCyz'  -\- p,;,  GzyVss' 

+  Pt/z  (Cyy'C^^'  +  CyzVzy')  +  ^««  (Csy'CaJs' 

+  C^s'CojyO  +  p^{GxyVyz'  +  Caj^'Cyy') ; 

^^jg,  =  j?a«Caj3'Ca?aj'  +  pyyQyz'Qyx'  4-  ^^^^  C^^'Cskb' 

+  i?2,.(C.y2'Csaj'  +  Qzz'Qy^x^)^  jpzJ^Cizz'Qxx'  ^-Qzx'Qxz') 
+i?«y  (CaJs'C^/a;' + Qxx'Q,yz') ; 

i^a/i/  =  ^a«Ca;a?'Ca^'  +  PyyCyx'Cyy'  +  jp^S^zx'Qzy' 

+Jpyz{Oyx'Qzy' + Qyy'Gzx')  -{-p^JS^zx' Cxy' + C^y'Caraj') 

+ _^a:y(Ca?a?'Cyy ' + Cxy'Cyx'). 

It  may  be  shown  that  for  every  state  of  stress  in  a  hody  there 
are  three  planes  perpendioular  to  each  other,  on  which  the  stress 
is  entirely  normal,    . 

[These  equations  are  useful  in  discussing  the  general  Theory 
of  the  Elasticity  of  Bodies?^ 

These  formulas  apply  to  oblique  axes  as  well  as  right,  only  it 
should  be  observed  when  they  are  oblique  that  jpyr^r  is  not  a 
stress  on  a  plane  normal  to  y'y  parallel  to  z\  but  on  a  plane  nor- 
mal to  x'  resolved  in  the  proper  direction. 


CHAPTER   YI 


VIRTUAL    VKLOCITIES. 


99,  Bef.  If  the  point  of  application  of  a  force  be  moved 
in  the  most  arbitrary  manner  an  indefinitely  small  amount,  the 

jirpjection  of  the  path  thus  described  on  the 
Anginal  actibn-line   of  the  force  is  called  a  h 

virtual  velocity.    The  product  of  the  force        /I ^ 

into  the  virtual  velocity  is  called  the  mrtual  ^^^  ^ 

moment.    Thus,  in  Fig.  87,  if  a  be  the  point  of 

application  of  the  force  i^  and  ab  the  arbitrary  displacement, 

ac  will  be  the  virtual  velocity,  and  Kao  the  virtual  moment. 
The  path  of  the  displacement  must  be  so  short  that  it  may  be 

considered  a  straight  line ;  but  in  some  cases  its  length  may  be 

finite. 

If  the  projection  falls  upon  the  action-line,  as  in  Fig.  87,  the 

virtual  velocity  will  be  considered  positive,  but  if  on  the  line 

prolonged,  it  will  be  negative. 

100.  Prop.  If  several  concurring  forces  are  in  equilihHum,, 
the  algebraic  sum  of  their  virtual  moments  will  be  zero. 

Using  the  notation  of  Article  (47),  and  in  addition  thereto 
let 

I  be  the  length  of  the  displacement ;  and 
j>,  q,  and  r  the  angles  which  it  makes  with  the  respective 
coordinate  axes; 

then  will  the  projections  of  I  on  the  axes  be 

I  cos  J?,        I  cos  q,        I  cos  r, 

respectively.      Multiplying  equations  (50),  by   these   respect- 
ively, we  have 

FiG09>  cLil  co&jp  +  7^  cos  flj  Z  cos^  +  etc.  =  0 ; 
i^  cos  ^j  ?  cos  ^  +  7^  cos  ^2  ^  cos  2"  -^  etc.  =  0 ; 
^  cos  7i  Z  cos  r  -f-  i^  cos  72  Z  cos  /♦  +  etc.  =  0. 


160  VIRTUAL  VELOCITIES.  [101.  J 

Adding  these  together  term  by  term,  obsej-viiig  that 
cos  a  cos^  +  cos  ^  cos  q  +  cos  7  cos  r  =  cos  {F.I) ; 

which  is  the  cosine  of  the  angle  between  the  action -line  of  F 
and  the  line  Z;  and  that  I  cos  {Fl)  =  hf  (read,  variation  f) 
is  the  virtual  velocity  of  F^  we  have 

F,8f,  +  i^S/2  +  i^3?/3  +  etc.  =  XFhf  =  0 ;       (116) 
which  was  to  be  proved. 

101.  If  <^^y  number  of  forces  in  a  system  are  in  equilih^ 
rium^  the  sum  of  their  virtual  moments  will  he  zero. 

Conceive  that  the  point  of  application  of  each  force  is  con- 
nected with  all  the  others  by  rigid  right  lines,  so  that  the  action 
of  all  the  forces  will  be  the  same 
as  in  the  actual  problem.  If  any 
of  the  lines  thus  introduced  are 
not  subjected  to  stress,  they  do 
not  form  an  essential  part  of  the 
system  and  may  be  cancelled  at 
first,  or  considered  as  not  having 
been  introduced.  Let  the  system 
receive  a  displacement  of  the 
most    arbitrary   kind.     At    each  ^^^-  ^' 

point  of  application  of  a  force  or  forces,  the  stresses  in  the  rigid 
lines  which  meet  at  that  point,  combined  with  the  applied 
force  or  forces  at  the  same  point,  are  necessarily  in  equilibrium, 
and  by  separating  it  from  the  rest  of  the  system,  we  have  a 
system  of  concurrent  forces.  Hence,  for  the  point  B^  for  in- 
stance, we  have,  according  to  (116), 
Flf-\-  i^a/i+  etc.  +  BChBC+BAhBA  +  BDIBD  =  0; 

in  which  BO^  etc.,  are  used  for  the  tension  or  compression 
which  may  exist  in  the  line.  But  when  the  point  C  is  consid- 
ered, we  will  have  B  CSB  C  with  a  contrary  sign  from  that  in 
the  preceding  expression,  and  hence  their  sum  will  be  zero. 

Proceeding  in  this  way,  as  many  equations  may  be  estab- 
lished as  there  are  points  of  application  of  the  forces;  and 
adding  the  equations  together,  observing  that  all  the  expressions 
which  represent  stresses  on  the  lines  disappear,  we  finally  have 

XFhf=0.  (117) 


[101.]  EXAMPLES.  161 

The  converse  is  evidently  true,  that  when  the  sum  of  the  vw* 
tual  moments  is  zero  the  system  is  in  equilibrium. 

Equations  (116)  and  (llT)  are  no  more  than  the  vanishing 
equations  for  work.  If  a  system  of  forces  is  in  equilibrium  it 
does  no  work.  This  principle  is  easily  extended  to  Dynamics. 
For,  the  work  which  is  stored  in  a  moving  body  equals  that 
done  by  the  impelling  force  above  that  which  it  constantly  does 
in  overcoming  resistances.  Thus,  when  friction  is  overcome, 
the  impelling  forces  accomplish  work  in  overcoming  this  resist- 
ance, and  all  above  that  is  stored  in  the  moving  mass.  Letting 
R  be  the  resultant  of  all  the  impressed  forces  producing 
motion,  and  s  the  path  described  by  the  body,  we  have 

Ehr-  Sm^Ss  =  0.  (118) 

This  is  the  most  general  principle  of  Mechanics,  and  M. 
Lagrange  made  it  the  fundamental  principle  of  his  celebrated 
work  on  Mecanique  Analytique,  wliich  consisted  chiefly  of  a 
discussion  of  equation  (118). 

Examples. 

1.  Determine  the  conditions  of  equilibrium  of  the  straight 
lever. 

Let  AB  be  the  lever,  having 
a  weight  P  at  one  end  and  TTat 
the  other,  in  equilibrium  on  the 
fulcrum  G. 

Conceive  the  lever  to  be 
turned  infinitesimally  about   G,  fio.  89. 

taking  the  position  CJD,  then  will  Aa,  which  is  the  projec- 
tion of  the  path  AC  on  the  action-line  of  P,  be  the  virtual 
velocity  of  P ;  and  similarly  Bh  will  be  the  virtual  velocity  of 
W,    The  former  will  be  positive  and  the  latter  negative;  hence 

P.Aa  -  W.Bb=  0. 

The  triangles  Aa O  Q,r\d  AOG  at  the  limit  are  similar,  having 
the  right  angles  AaC  and  AOG,  a  AC  —  AGC,  and  the  re- 
maining angle  equal.      Similarly,  hDB  is  similar  to  BGD, 
11 


162 


VIRTUAIi  VELOCITIES. 


aoi.j 


•*•    m^-DG^^^^''^^^''^'^^BG'^ 

which,  substituted  in  the  preceding  expression,  gives 

F,AG  =  WBG\ 

that  is,  the  weights  are  inversely  proportioned  to  the  arms. 

If  the  lever  be  turned  about  the  end  A,  we  would  find  in 
a  similar  manner  that  {P  +  W).A  G  =  W.AB ;  in  which 
P  +  TFis  the  reaction  sustained  by  the  fulcrum  G, 

2.  Find  the  conditions  of 
equilibrium  of  the  bent  lever. 

Let  ^ 6^  and  GB  be  the  arms 
of  the  lever  and  G  the  fulcrum. 
Let  it  be  turned  slightly  about 
G ;  then  will  Aa  and  Bh  be  the 
respective  virtual  velocities  of 
Fio.  90.  JP  and  Tf; 

.-.  -  P.Aa  +   W.hB  =  0. 

From  G  draw  GO  perpendicular  to  PAy  then  will  the  tri- 
angle A  CG  be  similar  to  AaA',  having  the  angle  AaA'  =  A  OG; 
and  a  A  A'  =  OGA.  Similarly,  the  triangle  BDG  is  similar  to 
BhB'\ 


Aa 
'Bb 


GO 
GD' 


which,  combined  with  the  preceding  equation,  gives 

P.GC=  W,GD; 

that  is,  the  weights  are  inversely  proportional  to  their  horizon- 
tal distances  from  the  fidcrum. 

3.  Find  the  conditions  of  equilibrium  of  the  single 
pulley. 

In  Fig.  91,  let  the  weight  P  be  moved  a  distance 
equal  to  ah,  then  will  W  be  moved  a  distance  cd  = 
ah  ;  hence,  we  have 

-  P. ah  +  Wxd  =  0 ;    .-.  P  =  W^ 


r\ 


P      w 

Fio.  91. 


iioi.i 


EXAMPLES. 


163 


4r.  On   the   inclined  plane  AC,  a,  weight  P  is  held  by  a 
force  TT  acting  parallel  to  the  plane  ; 
required  the  relation  between  P  and 
W, 

de  =  ah  will  be  the  virtual  velocity 
of  Wj  and  ac  that  of  P ;  and  we  have 

-  P.OG  -f  W.ab  =  0. 
From  the  similar  triangles  abc  and 
ABC,  we  have  Pio.  92. 


P.CB=  W.AC;  or 


ac  __  CB 
ah  ~~  AC 

P:  W  ::  AC:  CB. 

6.  On  the  inclined  plane,  if  the 
weight  P  is  held  by  a  force  W, 
acting  horizontally,  reqfuired  the 
relation  between  P  and  W. 

The  movement  being  made,  cd 
will  be  the  virtual  velocity  of  W, 
which  at  the  limit  equals  ae, 
and  he  will  be  the  virtual  velo- 
city of  P,  and  we  have  Pio.  98. 

-  P.he  +  W.ae  ==  0 ;  B,nd  ae  :  eh  : :  AB  :  BC, 
.'.P.CB=:  W.BA; 
or,  the  weight  is  to  the  horizoiital  force  as  the  base  of  the  tri- 
angle is  to  its  altitude 

6.  In  Fig.  27  show  that  Pdr  — 

7.  One  end  of  a  beam  rests 
on  a  horizontal  plane,  and  the 
other  on  an  inclined  plane ;  re- 
quired the  horizontal  pressure 
against  the  inclined  plane. 

This   involves  the  principle 
of  the  wedge ;   for  the  block 
AB  C  may  represent  one-half  of 
a  wedge  being  forced  against  the  resistance  W.     Conceive  the 
plane  to  be  moved  a  distance  AA' ,  and  that  the  beam  turns 


^Ydy, 


Fio.  94. 


164  VIRTUAL  VELOCITIES.  [lOl.J 

about  the  end  Z>,  but  is  prevented  from  sliding  on  the  plane ; 
tlieu  will  the  virtual  velocity  of  the  horizontal  pressure  be  AA ', 
and  that  of  the  weight  will  be  & ;  hence,  for  equilibrium  we 
have 

W,Eg  -  I^AA'  =  0.  (a) 

We  now  find  the  relation  between  J^c  and  AA\ 
Let  I  =  Z>i^  the  length  of  the  beam  ; 

a  =  DA\  the  distance  from  D  to  the  centre  of  gravity  of 

the  beam ; 
a=  CAB;        p  =  ADK  *' 

The  end  at  F  will  describe  an  arc  FF'  about  Z>  as  a  centre. 
From  F'  draw  F'd  parallel  to  AA\  and  from  F  drop  a  per- 
pendicular Fe  upon  dF\  Then,  from  similar  triangles,  we 
have 

Fe  =  -  Fg, 
a 

FF'  will  be  perpendicular  to  DF,  and  Fe  perpendicular  tc 
dF\  hence 

eFF'  =  AI)F=  ^',  dFe=:  90°—  a; 

.-.  dFF'=  90°-a  +  /5; 


and 


FF'  =  FeseoB  =  -  Eg  sec  /3. 


The  triangle  FdF'  gives 

FF'  sin  a 


AA  =  dF'       sin  (90^  _  a  +  /S)' 
hence, 

Eg   ^a    sin  a  cos  ^  ^ 
AA'  ""  1    cos  (a  -  yS) ' 

which,  substituted  in  equation  («)  above,  gives 

a    sin  a  cos  /S 


P=  TT 


Z    cos  (a  —  y3)' 


8.  Deduce  the  formula  for  the  triangle  of  forces  from  the 
principle  of  Virtual  Velocities. 


CHAPTEE    V^II. 

MOMENT   OF   INERTIA. 

(This  chapter  may  be  omitted  until  its  principles  are  needed  hereafter  (sec 
Ch.  X)  Although  the  expression  given  below,  called  the  Moment  of  Inertia^ 
comes  directly  from  the  solution  of  certain  mechanical  problems,  yet  its  prin- 
ciples may  be  discussed  without  involving  the  idea  of  force^  the  same  as  any 
other  mathematical  expression.  The  term  probably  originated  from  the  idea 
that  inertia  was  considered  a  forcCy  and  in  most  mechanical  problems  which 
give  rise  to  the  expi'ession  the  moment  of  a  force  is  involved.  But  the  expres- 
sion is  not  in  tho/orm  of  a  simple  moment.  If  we  consider  a  moment  as  the 
product  of  a  quantity  by  an  arm,  it  is  of  the  form  of  a  moment  of  a  moment. 
Thus,  dA  being  the  quantity,  ydA  would  be  a  moment,  then  considering  this 
as  a  new  quantity,  multiplying  it  by  y  gives  y'^dA^  which  would  be  a  moifient 
of  the  moment.  Since  we  do  not  consider  inertia  as  a  force,  and  since  all 
these  problems  may  be  reduced  to  the  consideration  of  geometrical  magnitudes, 
it  appears  that  some  other  term  might  be  more  appropriate.  It  being,  how- 
ever, imiversally  used,  a  change  is  undesirable  unless  a  new  and  better  one  be 
universaUy  adopted.) 

DEFINITIONS. 

102.  The  expression, /^c?^,  in  which  dA  represents  an  ele- 
ment of  a  body,  and  y  its  ordinate  from  an  axis,  occurs  fre- 
quently in  the  analysis  of  a  certain  class  of  problems,  and  hence 
it  has  been  found  convenient  to  give  it  a  special  name.  It  is 
called  the  moment  of  inertia, 

THE  MOMENT   OF   INERTIA   OF   A  BODY 

is  the  sum,  of  thejproducts  ohtained  hy  multiplying  each  element 
of  the  hody  hy  the  square  of  its  distance  fror^x  an  axis. 

The  axis  is  any  straight  line  in  space  from  which  the  ordinate 
is  measured. 

The  quantity  dA  may  represent  an  element  of  a  line  (straight 
or  curved),  a  surface  (plane  or  curved),  a  volume,  weight,  or 
mass ;  and  hence  the  above  definition  answers  for  all  these 
quantities. 


166 


MOMENT  OF  INERTIA. 


[103.J 


The  moment  of  inertia  of  a  plane  surface,  when  the  axis  lies 
in  it,  is  called  a  rectangular  moment;  but  when  the  axis  is 
perpendicular  to  the  surface  it  is  called  ^jpolar  moment. 


103.  Examples. 

1.  Find  the  moment  of  inertia  of  a  rect- 
angle in  reference  to  one  end  as  an  axis. 

Let  h  =  the  breadth,  and  d  =  the  depth 
of  tlie  rectangle.  Take  the  origin  of  coor- 
dinates at  0, 

We  have  dA  =  dt/dx ; 

i    fdydx  =zh   I    fdy  =  ibd^. 


Fio  95. 


and 


/  = 


Fig,  «6. 


2.  What  is  the  moment  of  inertia  of  a 
rectanojle  in  reference  to  an  axis  through 
the  centre  and  parallel  to  one  end  ? 

'-^  A71S.  ^.^hd^, 

3.  What  is  the  moment  of  inertia  of  a 
straight  line  in  reference  to  an  axis  through 

one  end  and  perpendicular  to  it,  the  section  of  the  line  being 

considered  unity  ? 

Am.  i^. 

'U.  Find  the  moment  of  inertia  of  a  circle  in  reference  to  an 
axis  through  its  centre  and  perpendicular  to  its  surface. 
We  represent  the  j^olar  moment  of  inertia  by  I^, 

Let       r  =  the  radius  of  the  circle ; 

p  =  the  radius  vector  ;  ' 

6  =  the  variable  angle  ;  then 
dp  =  one  side  of  an  elementary  rectangle ; 
pdd  ==  the  other  side ;  and 
dA.  =  pdpdd ; 

and,  according  to  the  definition,  we  have 

•2t 


h  = 


p'dpde  =  im*. 


1 104.  J  EXAMPLES.  167 

5.  What  is  the  moment  of  inertia  of  a  circle  in  reference 
to  a  diameter  as  an  axis  ?     (SeeArticle  105.) 

Ans.  iirr*. 

6.  What  is  the  moment  of  inertia  of  an  ellipse 
in  reference  to  its  major  axis ;  a  being  its  semi- 
major  axis  and  i,its  semi-minor?  fio-  97. 

Ans.  ^TraP, 

7.  Find  the  moment  of  inertia  of  a  triangle  in  reference  to 
an  axis  through  its  vertex  and  parallel  to  its  base. 

Let  b  be  the  base  of  the  triangle,  d  its  altitude,  and  x  any 
width  parallel  to  the  base  at  a  point  whose  ordinate  is  y ;  then 
dA  =  dosdi/,  and  we  have 


^=    /       /        y^dydx=r.-^  I     fdy  =  \hd^. 

8.  What  is  the  moment  of  inertia  of  a  triangle  in  reference 

to  an  axis  passing  through  its  centre  and  parallel  to  the  base  ? 

Ans,  -^^MK 
*         

9.  What  is  the  moment  of  inertia  of  an  isosceles  triangle  in 

reference  to  its  axis  of  symmetry  ? 

Ajis.  -^b^d. 

10.  Find  the  moment  of  inertia  of  a  sphere  in  reference  to  a 
diameter  as  an  axis. 

The  equation  of  the  sphere  will  \)Q  s^  +  y^  +  £^  =  I^.  The 
moment  of  inertia  of  any  section  perpendicular  to  the  axis  of  x 
will  be  Jtt^  ;  hence  for  the  sphere  we  have 

/=    f^  ^y'dx='Tr   f    {B^-x^fdx  =  -^irI^. 

FORMULA    OF   REDUCTION. 

104.  The  moinent  of  inertia  of  a  body,  in  reference  to  any 
axis,  equals  the  moinent  of  inertia  in  reference  to  a  parallel 
axis  jpassing  through  the  centre  of  the  body  plus  the  product 
of  the  area  {or  volume  or  7nass)  by  the  square  of  the  distance 
between  the  axes. 


168 


FORMULA  OF  REDUCTION. 


[104.J 


This  proposition  for  plane  areas  was  proved  in  Article  80. 

To  piove  it  generally,  let  Fig.  98  re- 
present the  projection  of  a  body  upon  the 
plane  of  the  papei*,  B  the  projection  of 
an  axis  passing  through  the  centre  of  the 
body,  A  any  axis  parallel  to  it,  C  the 
projection  of  any  element;  AC  —  r,  BO 
=  ri,  the  angle  CBJ)  =  6,  and  F  =  the 
volume  of  the  body. 
Then 


J^  =  fr^d  V  will  be  the  moment  of  inertia  of  the  volume  in 
reference  to  the  axis  through  the  centre ;  and 

I  —Jr^dV,  the  moment  in  reference  to  the  axis  through  A. 

Let  AB  -:  Z>,  then  Ai:  =  I)  -{-  r,  cos  6,  and  r^  =  r}  sin^  6 
+  (i>  +  n  cos  ej ; 

,\  fM  V  =frM  V  +  2I)/h  cos  edV+  Dyd  V, 

But  fr^  cos  6dV  =  0,  since  it  is  the  statical  moment  of  the 
body  in  reference  to  a  plane  perpendicular  to  AD  passing 
through  the  centre  of  the  body  and  perpendicular  to  the  plane 
of  the  paper,  therefore  the  preceding  equation  becomes 

/=/, -F-  Fi>^;  (119) 

which  is  called  the  formula  of  reduction. 
From  this,  we  have 

7i  =  I-  VJD^.  (120) 


Examples. 

1.  The  moment  of  inertia  of  a  rectangle  in  reference  to  one 
end  as  an  axis  being  ^hd^^  required  the  moment  in  reference 
to  a  parallel  axis  through  the  centre. 

Equation  (120)  gives 

7,  =  ^hd^  -  hd  (idf  =  -^M\ 

2.  Given  the  moment  of  inertia  of  a  triangle  in  reference  to 
an  axis  through  its  vertex  and  parallel  to  the  base,  to  find  the 
moment  relative  to  a  parallel  axis  through  its  centre. 


I105.J 


EXAMPLES. 


169 


Example  7  of  the  preceding  Article  gives  /=  ihd^;  hence 
equation  (120)  gives 

I,  =  ibd'  -  ibd  iidf  =  ^hd^ 

3.  Find  the  moment  of  the  same  triangle  in  reference  to  the 
base  as  an  axis. 
Equation  (119)  gives 

/=  i,M^  +  ibd  iid)^  =  -^y^hdK 

105,  To  FIND  THE  RELATION  BETWEEN  THE  MOMENTS  OF  IN- 
ERTIA IN  REFERENCE  TO  DIFFERENT  PAIRS  OF  RECTANGULAR 
AXES    HAVING    THE    SAME    ORIGIN. 

Let  X  and  y  be  rectangular  axes, 
a?i  and  yi,  also  rectangular, 
having  the  same  origin  ; 
a  =  the    angle   between  x 

and  a?i ; 
I^  =  the  moment  of  inertia 
relatively  to  the  axis 
Xj  similarly  for 

B  =  JxydA  ;  and 

For  the  transformation  of  coordinates  we  have 

a?!  =  y  sin  a  -I-  a?  cos  a ; 
yi  =  y  cos  a  —  0?  sin  a ; 

x}  +  y^  =1  a?  +  y^. 


Fio.  99. 


Also 


Hence, 


dA  =  dxdy  =  dx^dy'^. 


Ix^  =  J''y\  dA  =  Txiios^  a  -^  ly  sin*  a  —2^5  cos  a  sin  a ; 

Jy^  =  I^  sin^  a  -^  ly  cos'^  a  +  ^B  cos  a  sin  a ; 

B^  =  {Tx  —  ii )  cos  a  sin  a  +  ^  (cos^  a  —  sin^  a) ; 

.*•    Ixi    "r    J-tfi    -^  -^x      I     •'y    ^^  -fp  > 


(121) 


the  last  value  of  which  is  found  from  the  expression  fy^dA  + 
fx^dA  =  f{y^  +  x^)  dA  =ff?dA  =  Ij,;  which  shows  that  the 


170  MOMENT  OF  INERTIA.  [105.] 

polar  moment  equals  the  sum  of  two  rectangular  moments,  the 
origin  being  the  same.  If  the  rectangular  moments  equal  one 
another,  we  have  Ip  =  2Jx .  Thus,  in  the  circle,  Ip  =  iirr*. 
(See  Ex.  4,  x\rticle  103),  hence  I^  =  iirr^ 

The  last  of  equations  (121)  is  an  isotropic  function ;  since 
the  sum  of  the  moments  relatively  to  a  pair  of  rectangular 
axes,equals  the  sum  of  the  moments  relatively  to  any  other  pair 
of  rectangular  axes  having  the  same  origin ;  or,  in  other  words, 
the  sum  of  the  moments  of  inertia  relatively  to  a  pair  of  rect- 
angular axes,  is  constant. 

To  find  the  maximum  or  minimum  moments  we  have,  from 
the  preceding  equations, 


da 

=  -(/. 

—  ly)  COS  a  sin  a  - 

-  B  (cos2  a  - 

■  sin^ 

a)  = 

0; 

and 

da 

=  +(/»- 

-  i^ )  cos  a  sin  a  + 
/.  J?i  =  0. 

B  (cos^  a  - 

sin^ 

a)  = 

0; 

From  the  first  < 

lY  second  of  these  we  have 

—iB 
I.- 

2  cos  a  sin  a 
ly  ~~  cos^  a  —  sin^  < 

-  =  tan  2a. 

It  may  be  shown  by  the  ordinary  tests  that  when  Ix^  is  a 
maximum,  ly^  will  be  a  minimum,  and  the  reverse ;  hence 
there  is  always  a  pair  of  rectangular  axes  in  reference  to  one 
of  which  the  moment  of  inertia  is  greater  than  for  any  other 
axis,  and  for  the  other  it  is  less. 

These  are  called  principal  axes. 

Thus,  in  the  case  of  a  rectangle,  if  the  axes  are  parallel  to 
the  sides  and  pass  through  the  centre,  we  find 


1/3 


yd  A  =  0 ; 
id 

hence  x  and  y  are  the  axes  for  maximum  and  minirauiQ 
moments ;  and  if  <i  >  Z»,  j)^bd^  is  the  maximum,  and  -^jh^d  a 
minimum  moment  of  inertia  for  all  axes  passing  through  the 
origin.     In  a  similar  way  we  find  that  if  the  origin  be  at  any 


P06.] 


EXAMPLES. 


171 


other  point  the  axes  must  be  parallel  to  the  sides  for  maximum 
and  minimum  moments. 

The  preceding  analysis  gives  the  position  of  the  axes  for 
maximum  and  minimum  moments,  when  the  moments  are 
known  in  reference  to  any  pair  of  rectangular  axes.  But  if  tlie 
axes  for  maximum  and  minimum  moments  are  known  as  I^  and 
ly^  then  j5  =  0;  and  calling  these  ^  and  i^,  Eqs.  (121) 
become 

I^^zzz  I^  cos^  a  +  Iy>  sln^  a ;  \ 

I^  sin2  a  +  ly,  cos2  a ;  I  (122) 

(/a/  —  ly*)  COS  a  sin  a.   ) 

In  .the  case  of  a  square  when  the  axes  pass  tlirough  the 
centre  I^'  —  ly' \ 

.*.  I^^  =  I^  (cos^  a  -f  sin^  a)  =  j^ ; 
ly^  =  lyf,  and 

hence  the  moment  of  inertia  of  a  square  is  the  same  in  refer- 
ence to  all  axes  passing  through  its  centre.  The  same  is  true 
for  all  regular  polygons,  and  hence  for  the  circle. 


Examples. 


1.  To  find  the  moment  of  inertia  of  a  rect- 
angle in  reference  to  an  axis  through  its  cen- 
tre and  inclined  at  an  angle  a  to  one  side,  we 
have 

I^  =  -^^hd^  and  ly  =  -^^Wd 
.*.  7a;j  =  tV^<^  {d"^  cos^  a  +  5^  sin^  a) ; 

-4i  =  A^<^  (<^^  siii^  a  •\-  ¥  cos^  a).  fig.  lOO. 

2.  To  find  the  moment  of  inertia  of  an  isosceles  triangle  in 
reference  to  an  axis  through  its  centre  and  inclined  at  an  angle 
a  to  its  axis  of  symmetry. 

We  have  I^  —  ^hd^  and  ly  —  -^^V^d,  in  which  h  is  the  base 
and  d  the  altitude ; 


^^Id  {d'^  cos2  a  ■\-  ily"  sin'  a) 
j>-o56?(c?=^sin2a  -f  Wao^^  a). 

The  moment  of  inertia  of  a  regular  polygon  about  an  axis 


/,  = 


172 


MOMENT  OF  INERTIA. 


[100.J 


through  its  centre  may  be  found  by  dividing  it  into  triangles 
having  their  vertices  at  the  centre  of  the  polygon,  and  for 
bases  the  sides  of  the  polygon ;  then  finding  the  moments  of 
the  triangles  about  an  axis  through  their  centre  and  parallel  to 
the  given  axis  and  reducing  them  to  the  given  axis  by  the  for- 
mula of  reduction. 

If  R  be  the  radius  of  the  circumscribed  circle,  r  that  of  the 
inscribed  circle,  and  A  the  area  of  the  polygon ;  then,  for  a 

A B  regular  polygon,  we  would  find  that 

/=  ^A(m  +  2/^). 
For  the  circle  R  =  r^ 

as  before  found. 
For  the  square,  /»  =  JJ,  i?  =  ^5 1^2,  and  ^  =  5*; 

as  before  found. 


Fro.  102. 


106.  Examples  of  the  moment  of  inertia  of  solids. 

(The  following  results  are  taken  from  Mosley's  Mechanics  and  Engineering.) 

1.  The  moment  of  inertia  of  a  solid  cylinder 
about  its  axis  of  symmetry,  r  being  its  radius  and 
h  its  height,  is  iirhr^,  ^  -^      ( 

2.  If  the  cylinder  is  hollow,  g  the  thickness  of 
the  solid  part  and  R  the  mean  radius  (equal  to 
one-half  the  sum  of  the  external  and  internal  radii) 
then  /=  27rAc^  {R"  +  Jo^). 

3.  The  moment  of  inertia  of  a  cylinder  in  reference  to  an 
axis  passing  through  its  centre  and  perpendicular  to  its  axis  of 
symmetry  is  ^irhr^  {i^  +  -JA^). 

4.  The  moment  of  inertia  of  a  rectangular  paral- 
lelopipedon  about  an  axis  passing  through  its  cen- 
tre and  parallel  to  one  of  its  edges.  Let  a  be  the 
length  of  the  edge  parallel  to  the  axis,  and  b  and  o 
the  lengths  of  the  other  edges,  then  T  =  ^^  abo 
Q)^  +  <^)  z=z  ^  of  the  volume  multvplied  hy  the 
Pio.  103.       square  of  the  diagonal  of  the  hose. 


[107.] 


RADIUS  OF  GYRATION. 


173 


5.  The  moment  of  inertia  of  an  upright   triangular  prism 
having  an  isosceles  triangle  for  its  base,  in  reference 
to    a   vertical    axis  passing   through   its   centre    of 
gravity. 

Let  the  base  of  the  triangle  be  «,  its  altitude  h,  and 
the  altitude  of  the  prism  be  A,  then 

Via.  104. 

6.  The  moment  of  inertia  of  a  cone  in  reference  to  an  axia 
of  symmetry  is  ^irr^L  {r  being  the  radius  of  the  base  and  h 
the  altitude.) 


Fio.  105.  Fio.  106.  Fia.  107. 

7.  The  moment  of  inertia  of  a  cone  in  reference  to  an  axis 
through  its  centre  and  perpendicular  to  its  axis  of  symmetry  is 

8.  The  moment  of  inertia  of  a  sphere  about  one  of  its  diam- 
eters is  -^irl^. 

9.  The  moment  of  inertia  of  a  segment  of  a  sphere  about  a 
diameter  parallel  to  the  plane  of  section. 

Let  R  be  the  radius  of  the  sphere,  and  h 
the  distance  of  the  plane  section  from  the 
centre,  then 

/  =  ^TT  (1 6i?«  +  157^^  +  10i?2J8  -  95«).  FIO.  108. 


sphere  about  i 


RADIUS   OF   GYRATION. 

107,  We  may  conceive  the  mass  to  be  concentrated  at  such 
a  point  that  the  moment  of  inertia  in  reference  to  any  axis  will 
be  the  same  as  for  the  distributed  mass  in  reference  to  the  same 
axis. 

The  radius  of  gyration  is  the  distance  from  the  moment 
axis  to  a  point  in  which,  if  the  entire  mass  be  concentrated,  the 
moment  of  inertia  will  be  the  same  as  for  the  distributed  mass 


174  EXAMPLES.  [107.] 

The  jprincvpal  radius  of  gyration  is  the  radius  of  gyration 
in  reference  to  a  moment  axis  through  the  centre  of  the  mass. 
Let  Tc  =  the  radius  of  gyration  ; 

h^  =  the  principal  radius  of  gyration  ; 
M  =  the  mass  of  the  body ;  and 
D  =  the  distance  between  parallel  axes ; 
then,  according  to  the  definitions  and  equation  (119),  we  have 
M^  =  Sm?^ 

=  Xmr^  +  MIJ^ 

/.   W=h}  -\-  ]J*^  (123) 

from  which  it  appears  that  ^  is  a  minimum,  for  1?  =  0,  in  which 
case  h  —  \\  that  is,  the  jprindjpal  radius  of  gyration  is  the 
minimum  radius  for  jparallel  axes. 
We  have 

hence,  the  square  of  the  principal  radius  of  gyration  equals  the 
mbment  of  inertia  in  reference  to  a  moment  axis  through  the 
centre  of  the  body  divided  by  the  mass. 

Examples. 

1.  Find  the  principal  radius  of  gyration  of  a  circle  in  refer- 
ence to  a  rectangular  axis. 

Example  5  of  Article  103  gives,  Ii  =  iTr/'*,  which  is  the 
moment  of  an  area,  hence,  we  use  tt/^  for  J/,  and  have 

2.  For  a  circle  in  reference  to  a  polar  axis,  ki^  —  ^. 

3.  For  a  straight  line  in  reference  to  a  moment  axis  perpen- 
dicular to  it,  k^  =  -^jP. 

4.  For  a  sphere,  ^^  =  fr*. 

5.  For  a  rectangle  whose  sides  are  respectively  a  and  J,  in 
reference  to  an  axis  perpendicular  to  its  plane,  Jc^—^^  {a^+¥). 

6.  Find  the  principal  radius  of  gyration  of  a  cone  when  the 
moment  axis  is  the  axis  of  symmetry. 


CHAPTER   YIII. 


MOnON   OF   A   PARTICLE   FREE   TO   MOVE   IN   ANT   DIRECTION. 


108.  A  free,  material  particle,  acted  upon  by  a  system  of 
forces  which  are  not  in  equilibrium  among  themselves,  will 
describe  a  path  the  direction  of  motion  in  which  immediately 
after  passing  any  point  will  depend  upon  its  direction  when 
it  arrives  at  the  point  and  the  resultant  of  the  forces  acting 
upon  it  at  the  point,  but  the  direction  of  the  acceleration  will 
be  in  the  direction  of  the  resultant  of  the  impressed  forces. 

Let  R  be  the  resultant  of  the  forces  acting  upon  a  particle 
whose  mass  is  m  at  the  point  whose  coordinates  are  a?,  y,  2,  and 
ds  the  path  wliich  would  be  described  by  the  effect  of  R  only ; 
then,  according  to  Article  21,  we  have 

Let  a  be  the  angle  between  the  action-line  of  the  resultant 
R  (or  of  the  arc  da)  and  the  axis  of  x\  multiplying  by  cos  a, 
we  have 

It  cos  a  —  m-T-^  cos  a  =  0 ; 

in  which  R  cos  a  is  the  x-coirvponent  of  the  resultant,  and 
according  to  equation  (51)  equals  X\  or,  in  other  words,  it  is 
the  projection  of  the  line  representing  the  resultant  on  the 
axis  of  a? ;  d?8  cos  a  is  the  projection  of  d^a  on  the  axis  of  oj,  and 
is  d^x.     Hence,  the  equation  becomes, 

-rr  d^x       _ 

r--J  =  0;     \  (124) 

7  ^'^       o 


and  similarly, 


176  MOTION  OF  A  [109.J 

which  are  the  equations  for  the  motion  of  a  particle  along  the 
coordinate  axes ;  and  are  also  the  equations  for  the  motion  of 
a  body  of  finite  size  when  the  action -line  of  the  resultant  passes 
through  the  centre  of  the  mass.  They  are  also  the  equations 
of  translation  of  the  centre  of  any  free  mass  when  the  forces 
produce  both  rotation  and  translation ;  in  which  case  m  should 
be  changed  to  M  to  represent  the  total  mass.     See  Article  38. 

VELOCITY   AND   LIVING   FORCE. 

109,  Multiplying  the  first  of  equations  (124)  by  dx,  the 
second  by  dy^  and  the  third  by  dz^  adding  and  reducing,  give 

Xdx  +  Ydy  +  Zdz  =  ^mdl -^ —\  =  ^md  -^ ; 

and  integrating  gives 

-      Hxdx  +  Ydy  +  Zdz)  =  im  ^'  =  ^^yiv^  +  O. 

The  first  member  is  the  work  done  by  the  impressed  forces  ; 
for  if  jS  be  the  resultant,  and  s  the  path,  then,  according  to 
Article  25,  equation  (26),  the  work  will  be  f  Rds^  and  by  pro- 
jecting this  on  the  coordinate  axes  and  taking  their  sum,  we 
have  the  above  expression.  The  second  member  is  the  stored 
energy  plus  a  constant. 

Let  X,  y,  Z  be  known  functions  of  a?,  y,  s,  and  that  the 
terms  are  integrable.  (It  may  be  shown  that  they  are  always 
integrable  when  the  forces  act  towards  or  from  fixed  centres.) 
Performing  the  integration  between  the  limits  oJq,  y^,  z^^  and 
aji,  j^i,  2i,  we  have 

^  (^0^  2^0^  ^o)  -  ^  (^>  yij  %)  =  i^  (^0  —  -y^i) ;     (125) 

hence,  the  work  done  by  the  impressed  forces  upon  a  body  in 
passing  from  one  point  to  another  equals  the  difference  of  the 
living  forces  at  those  points.  It  also  appears  that  the  velocity 
at  tw^o  points  will  be  independent  of  the  path  described  ;  also, 
that,  when  the  body  arrives  at  the  initial  point,  it  will  have  the 
same  velocity  and  the  same  energy  that  it  previously  had  at 
that  point. 


[109.1 


FREE  PARTICLB. 


177 


Examples. 

\,lfa  hody  is  jprojected  into  space^  and  acted  upon  only  hy 
gravity  and  the  impulse  /  required  the  curve  described  hy  the 
projectile. 

Take  the  coordinate  plane  xy  in  the 
plane  of  the  forces,  x  horizontal  and  y 
vertical,  the  origin  being  at  the  point  from 
which  the  body  is  projected. 

Let  Tr=  the  weight  of  the  body  ; 

V  =  the  velocity  of  projection  ;  and 
a  —  BAX  =  the  angle  of  elevation 

at  which  the  projection  is  made. 
We  have, 

X=0;         Y=-mg;        Z=0; 
and  equations  (124)  become 


Fio.  loy. 


3  =  0; 


-9 


(£y 

df 


0. 


Integrating,  observing  that  v  cos  a  will  be  the  initial  velocity 
along  the  axis  of  a?,  and  v  sin  a  that  along  y,  we  have, 

dx 


dt~ 

V  cos  a ; 

dy 
l[t~ 

V  sin  a  - 

and  integrating  again,  observing  that  the  initial  spaces  are  zero, 
we  have, 

x^^  vt  cos  a ;  ^ 

y  ■=  vt  ^m  a  —  \g^,  ) 
Eliminating  t  from  these  equations,  gives 

g^ 


y  —  X  tan  a  — 


(*) 


%v^  C08^  a ' 

which  is  the  equation  of  the  common  parabola,  whose  axis  is 
parallel  to  the  axis  of  y 
12 


178  PROJECTILES.  riW.l 

Let  A  be  the  height  through  which  a  body  must  fall  to  acquire 
a  velocity  v,  then  t^  =  2gh,  and  the  equation  {b)  becomes, 

To  find  the  range  AE^ 
make  y  —  0  in  equation  (5),  and  we  find  a?  =  0,  and 

X  =  AE  =  4A  cos  a  sin  a  =  2A  sin  2a  ;  (6?) 

which  is  a  niaxiumm  for  a  =  45°.  The  range  will  be  the  same 
for  two  angles  of  elevation,  one  of  which  is  the  complement  of 
the  other. 

The  greatest  height, 
will  correspond  to  x  =  h  sin  2a,  which,  substituted  in  (J),  gives, 

h  sin^  a.  (e) 

The  velocity  at  the  end  of  the  time  t 
is, 

or  by  eliminating  t  by  means  of  the  first  of  equations  (a),  we 
have. 


^=\/"'-2^^'^""  +  2W- 


(?) 


TYio'  direction  of  motion  at  any  point 
is  found  by  differentiating  equation  (c),  and  making 

tan  ^  =  -^  =  tan  a  —  ^,  ^  ,    .  ih) 

dx  2Acos  fl  ^  ' 

At  the  highest  point  ^  —  0,  .'.  x  =  A  sin  2a,  as  before  found. 

For  a?  =  2A  sin  2a,  we  have, 

tan  6  =  —  tan  a, 

or  the  angle  at  the  end  of  the  range  is  the  supplement  of  -the 

angle  of  projection. 

2.  A  body  is  projected  at  an  angle  of  elevation  of  45°,  and 
has  a  range  of  1,000  feet ;  required  the  velocity  of  projection, 
the  time  of  flight,  and  the  parameter  of  the  parabola. 


[109.]  EXAMPLES.  179 

3.  What  must  be  the  angle  of  elevation  in  crder  that  the 
horizontal  range  may  equal  the  greatest  altitude  ?  What,  that 
it  may  equal  n  times  the  greatest  altitude  1 

4.  Find  the  velocity  and  the  angle  of  elevation  of  a  projectile, 
so  that  it  may  pass  through  the  points  whose  coordinates  are 
Xi  =  400  feet,  yi  =  50  feet,  a^  =  600  feet,  and  ^g  =  ^^  feet. 

5.  If  the  velocity  is  600  feet  per  second,  and  the  angle  of 
elevation  45  degrees ;  required  the  range,  the  greatest  elevation, 
the  velocity  at  the  highest  point,  the  direction  of  motion  6,000 
feet  from  the  point  of  projection,  and  tlie  velocity  at  that  point. 

6.  If  a  plane,  whose  angle  of  elevation  is  ^,  passes  through 
the  origin,  find  the  coordinates  of  the  point  C,  Fig.  109,  where 
the  projectile  passes  it. 

7.  In  the  preceding  problem,  if  i  is  an  angle  of  depression, 
find  the  coordinates. 

8.  Find  the  equation  of  the  path  when 
the  body  is  projected  horizontally. 

9.  If  a  body  is  projected  in  a  due  south- 
erly direction  at  an  angle  of  elevation  a, 
and  is  subjected  to  a  constant,  uniform, 
horizontal  pressure  in  a  due  easterly  direc- 
tion ;  required  the  equations  of  the  path,  neglecting  the  resist- 
ance of  the  air. 

We  have 

X  =  0 ;         Y=  —  mg ;        Z  =  JF'{si  constant). 
The  projection  of  the  path  on  the  plane  xy  will  be  a  para- 
bola, on  X2  also  a  parabola. 

10.  If  a  body  is  projected  into  the  air,  and  the  resistance  of 
the  air  varies  as  the  square  of  the  velocity ;  required  the  equa- 
tion of  the  curve. 

(The  final  integrals  for  this  problem  cannot  be  found.  Approximate  solu- 
tions have  been  made  for  the  purpose  of  determining  certain  laws  in  regard  to 
gunnery.  It  is  desirable  for  the  student  to  establish  the  equations  and  make 
the  first  steps  in  the  reduction. ) 

(The  remainder  of  this  chapter  may  be  omitted  without  detriment  to  what 
follows  it.  It,  however,  contains  an  interesting  topic  in  Mechanics,  and  is  of 
Tital  importance  in  Mathematical  Astronomy  and  Physics.) 


180 


CENTRAL 


[110,111.1 


CENTRAL   FORCES. 

110.  Central  forces  are  such  as  act  directly  towards  or  from 
a  point  caljed  a  centre.  Those  which  act  towards  the  centre 
are  called  attractive^  and  are  considered  negative,  while  those 
which  act  from  the  centi-e  are  repulsive^  and  are  considered 
positive.     The  centre  may  be  fixed  or  movable. 

The  line  from  the  centre  to  the  particle  is  called  a  radius 
vector.  The  path  of  a  body  under  the  action  of  central  forces 
is  called  an  orbit. 

The  forces  considered  in  Astronomy  and  many  of  those  in 
Physics,  are  central  forces. 

GENERAL   EQUATIONS. 

111,  Consider  the  force  as  attractive,  and  let  it  be  represented 
by  —  i^  y    ,F 

Take  the  coordinate  plane  xy  in  the  plane  of         "'"^^'^ 
the  orbit,  the  origin  being  at  the  centre  of  the 
force,  and  OP  =  /•,  the  radius  vector,  then 


i^cosa  =  —  F—\ 


r=-Fcos^ 


F^; 


and  the  first  two  of  equations  (124)  become 


atr  r 


(126) 


To  change  these  to  polar  coordinates,  first  modify  them  by 
multiplying  the  first  by  y  and  the  second  by  a?,  and  subtracting, 
and  we  have 


my 


d^ 
df 


fllS.]  ,  FOROBS.  181 

and  multiplying  the  first  by  x  and  the  second  by  y,  and  adding, 
we  have 

Let  Q  =  POM  =  the  variable  angle,  then 

x  =  r  cos  6^  y  =  ^  sin  6, 

and  differentiating  each  twice,  we  find 

d^x  =  {d?r  -  rd^)  cos  B  —  {2drd0  +  rd'^d)  sin  6 ; 
<Z2y  =  (dJir  -  rdd^)  sin  ^  +  {Mrdd  -{-  r^^^^)  cos  0 ; 
which  substituted  in  the  preceding  equations,  give 

d^r 


d}r        (de\^  F  ,,^^, 


which  may  be  put  under  the  form 

ld(  ^dd 
rdi 


(^f)  =  0.         _  (128) 


Equation  (127)  shows  that  the  acceleration  along  r  is  the 
force  on  a  unit  of  mass ;  and  (128)  shows  that  there  is  no 
acceleration  perpendicular  to  the  radius  vector. 

112.  Princvple  of  equal  areas. — Integrate  equation  (128), 
and  we  have 

^§.^;  (129) 

and  integrating  a  second  time,  we  have  / 

J'Md  ==  Ct ;  (130)    '' 

the  constant  of  integration  being  zero,  since  the  initial  values 
of  t  and  Q  are  both  zero.  But  from  Calculusy/^<^^  is  twice  the 
sectoral  area  POX\  hence  the  sectoral  area  swept  over  by  the 
radius  vector  increases  directly  as  the  time ;  and  equal  areas 
wiU  he  passed  over  m  equal  times. 


182  CENTRAL  [113.] 

Making  ^  =  1,  we  find  that  C  will  be  twice  the  sectoral  area 
passed  over  in  a  unit  of  time. 

The  converse  is  also  true,  that  if  the  areas  are  proportional 
to  the  times  the  force  will  he  central. 

For,  multiplying  the  first  of  (126)  by  y  and  the  second  by  x 
and  taking  their  difference,  we  have 

or, 

Xy-  Jaj=..0; 

which  is  the  equation  of  a  straight  line,  and  is  the  equation  of 
the  action-line  of  the  resultant,  and  since  it  has  no  absolute 
term  it  passes  through  the  origin. 

113.   To  find  the  equation  of  the  orhit, 
eliminate  dt  from  equations  (127)  and  (128).     For  the  sake  of 

simplifying  the  final  equation,  make  r  =  — ,  and  (129)  becomes 

i=4-  (^^^> 

Differentiating  and  reducing,  gives 

,            du  ^dudt 

dr=z z=  —  C/ 


u^ 

or,  ^- ^  r^ 

dt"  dd' 


dd 


the  first  member  of  which  is  the  velocity  in  the  direction  of 
the  radius. 

Differentiating  again,  gives 

-d^--^^''w 

Substituting  these  in  equation  (127)  and  at  the  same  time 
making  m  equal  to  unity,  since  the  unit,  is  arbitrary,  we  have 


[114]  FORCES.  1S3 

or, 

d^u  F 

which  is  the  diiferential  equation  of  the  orbit. 

"When  the  law  of  the  force  is  known,  the  value  of  F  may  be 
substituted,  and  the  equation  integrated,  and  the  orbit  be 
definitely  determined. 

Multiplying  by  du  and  integrating  the  first  two  terms,  we 
have 


114,   Given  the  equation  of  the  orbit ^  to  find  the  law  of  the 
force. 

From  equation  (132),  we  have 


F=  G-u^(^  +  ^.  (134) 


Another  expression  is  deduced  as  follows :  let 

jp  =  the  perpendicular  from  the  centre  on  the  tangent, 
then  from  Calculus  we  have 


v2  


'i^de^  T^d^ 


r  =  -j^- 


d^        dr"  +  r'd^F- 

1 

"^TTa-  (135) 


Differentiating,  gives 

dH 


^^  +  ^ 


> 


184 

CENTRAL 

[115.] 

and  dividing 
we  have 

by  y,  i 

jubstituting,  du  ■ 

=  -. 

Mr, 

and 

reducing, 

1 

dp        jd^u    ,      \ 

which,  combined  with 

equation  (134), 

gives 

(136) 

which  is  a  more  simple  formula  for  determining  the  law  of  the 
central  force. 

115.  To  determine  the  velocity  at  any  point  of  the  orbit: 
We  have 

_d8  _d8  d6  _ds  dO 
'^'"dt~'dtTd  ~dd'dt 

ds 

=  CvJ^  -7^  (from  equation  (131)  ) 

=  -.  (from  Dif.  Oal\  (137) 

Hence,  the  velocity  varies  inversely  as  the  perjpendicular 
from  the  centre  ujpon  the  tangent  to  the  orhit. 

Another  expression  is  found  by  substituting  the  value  of  p, 
equation  (135),  in  (137). 

Hence,  , 

^=  Osl-^'ruK  (138) 

Still  another  expression  may  be  found  by  substituting  equa- 
tion (138)  in  (133);  hence 

v'^  C,  +  2fF^  (139) 

=  (7i  -  ^fFdr, 

Since  F  is  a  function  of  r,  the  integral  of  this  equation 
gives  V  in  terms  of  r,  or  the  velocity  depends  directly  upon  the 


[116, 117.1  FORCES.  185 

distance  of  the  body  from  the  centre.  Hence,  the  velocity  at 
any  two  points  in  the  orbit  is  independent  of  the  path  between 
them,  the  law  of  the  force  remaining  the  some. 

116.  To  determine  the  time  of  describing  any  portion  of 
the  orbit. 

To  find  it  in  terms  of  r,  eliminate  dS  between  equations  (129) 
and  (133),  reduce  and  find 


which  integrated  gives  the  time. 

To  find  it  in  terms  of  the  angle,  we  have  from  (129) 


t^-f^dd; 


from  which  r  must  be  eliminated  by  means  of  the  equation  of 
the  orbit,  and  the  integration  performed  in  reference  to  6. 

117,  To  find  the  components  of  the  force  along  the  tarhgent 
and  normal. 

Let  T  =  the  tangential  component ; 

IV  =  the  normal  component ; 
and  resolving  them  parallel  to  x  and  y,  we  have 


d^x                    dx           dy 
dt^                     ds            ds' 

"^dt^-^-^ds-^^ds' 

Elimin£ 

I  ting  iV  gives 

d^x  ^            d}y  ^         ^, 
m-^  dx\-  m  -^  dy  ^  Tds; 

or 

da     ■       ds' 

d^8 

m  -T-s. 
d^ 

£2 


186  CENTRAL  FORCES.  >  1117.) 

Eliminating  T  gives  ^^  ^    ^^€  ^^^ 


,7-  dx  cPy  dy  d^x 


ds  df         ,  da  d^     f^jL   -^  c^tpJi\^j^jf 

~  df'ds  \  d?  ■/ 

1? 
=  m-  ;  (141) 

hence  the  component  of  the  force  in  the  direction  of  the  nor- 
mal is  dejpendent  entirely  upon  the  velocity  and  radius  of  cur- 
vature. This  is  called  the  centrifugal  force.  It  is  the  measure 
of  the  force  which  deflects  the  body  from  the  tangent.  The 
force  directed  towards  the  centre  is  called  centripetal. 

If  ft)  =  the  angular  velocity  described  by  the  radius  of  cur- 
vature, then,  V  =  pea,  and  equation  (141)  becomes 

JV  =  may'p.  (142) 


Examples. 

1.  If  a  body  on  a  smooth,  horizontal  plane  is  fastened  to  a 
point  in  the  plane  by  means  of  a  string,  what  will  be  the  num- 
ber of  revolutions  per  minute,  that  the  tension  of  the  string  may 
be  twice  the  weight  of  the  body. 

2.  A  body  whose  weight  is  ten  pounds,  revolves  in  a  hori- 
zontal circle  whose  radius  is  five  feet,  with  a  velocity  of  forty 
feet  per  second ;  required  the  tension  of  the  string  which  holds 
it.     (Use  equation  (141).) 

3.  Required  the  velocity  and  periodic  time  of  a  body  re- 
volving in  a  circle  at  a  distance  of  n  radii  from  the  earth's 
centre. 

The  weight  of  the  body  on  the  surface  being  mg,  at  the  dis- 

(r  \       m^Q 
—  1  =  — 2",  and  this  is  a 


[117.]  EXAMPLES.  187 

measure  of  the  force  at  that  distance.     (Use  equation  (141)  or 
(139).) 


^-"{?)*--'"-cii 


(This  is  substantially  the  problem  which  Sir  Isaac  Newton  used  to  prove 
the  law  of  Universal  Gravitation.     See  Whewell's  IndtLCtive  Sciences. ) 

4.  A  particle  in  projected  from  a  given  point  in  a  given 
direction  with  a  given  velocity^  and  moves  under  the  action  of 
a  force  which  varies  inversely  as  the  square  of  the  distance 
from  the  centre  ;  required  the  orbit. 

Let  fi  =  the  force  at  a  unit's  distance,  then 

F=  iMu\ 
and  equation  (132)  becomes 

d^u  u* 


or, 

d^ 


the  first  integral  of  which  becomes  by  reduction 


dd  = 


\/^-(«-i.r 


in  which  A  is  an  arbitrary  constant,  and  the  negative  vahie  of 
the  radical  is  used. 
Integrating  again,  making  Oq  the  arbitrary  constant,  we  have 


6  —  6q  =  cos 

which  by  reduction  gives 

1         n    /        AO^  \ 

~  =  ;:  =  -^a  (l  +  ^  COS  (5  -  e,)),  (a) 


188  ORBITS  OP  [117.] 

which  is  the  general  polar  equation  of  a  conic  section,  the 
origin  being  at  the  focus.  As  this  is  the  law  of  Universal 
Gravitation,  it  follows  that  the  orbits  of  the  planets  and  comets 
are  conic  sections  having  the  centre  of  the  sun  for  the  focus. 
In  equation  {a),  6q  is  the  angle  between  the  major  axis  and  a 

line  drawn  through  the  centre  of  the  force,  and is  the 

eccentricity  =  e ;  hence  the  equation  may  be  written 


u 


^^[l  +  e  cos  {6  -  ^o)).  {b) 


The  magnitude  and  position  of  the  orbit  will  be  determined 
from  the  constants  which  enter  the  equation,  and  these  are 
determined  by  knowing  the  position,  velocity,  and  direction  of 
motion  at  some  point  in  the  orbit. 

Draw  a  figure  to  represent  the  orbit,  and  make  a  tangent  to 
the  curve  at  a  point  which  we  will  consider  the  initial  point. 
Let  yS  be  the  angle  between  the  path  and  the  radius  vector  at 
the  initial  point,  Tq  the  initial  radius  vector,  and  Vq  the  initial 
velocity;  then  at  the  initial  point 


1       /,       ^         ,  r^        dr  du 


u=-,    6  =  0,    cot5  =  -^=-^,  (.) 


and  from  equation  (b) 


du       lie    .    ^ 


which,  combined  with  equation  (c),  gives 


-  =  —  ^  sin  6^, 

W 

From  equation  (137) 

a  = 

Fo?'o  sin  ^ ; 

(/) 

[117.]  THE  PLANETS.  189 

which,  substituted   in   equations   {d)  and   {e\  and   the  latter 
divided  by  the  former,  gives 

/)         y^n  sill  P  ^os  /S 
^^"  ^«  =  /.-  FoVosin^^- 

Squaring  equations  (</)  and  (^),  adding  and  reducing  by  equa* 
tion  (/),  give 


,^,_i2^^(i.s).   ,, 


Hence,  when 

Fo*  >  — ,    6  >  1,  and  the  orbit  is  a  hyperbola, 

F"o'  =  — ,     6  =  1,  and  the  orbit  is  a  parabola, 

2a 
Yq  <.  — ,     e  <.  1,  and  the  orbit  is  an  ellipse. 

or  (see  example  23,  page  34)  the  orbit  will  be  a  hyperbola,  a 
parabola,  or  an  ellipse,  according  as  the  velocity  of  projection 
is  greater  than,  equal  to,  or  less  than  the  velocity  from  infinity. 
As  the  result  of  a  large  number  of  observations  upon  the 
planets,  especially  upon  Mars,  Kepler  deduced  the  following 


1.  The  planets  describe  ellipses  of  which  the  Sun  occupies  a 
focus. 

2.  The  radius  vector  of  each  planet  passes  over  equal  areas 
in  equal  times. 

3.  The  squares  of  the  periodic  times  of  any  two  planets  are 
as  the  cubes  of  the  major  axes  of  their  orbits. 

The  first  of  these  is  proved  by  the  preceding  problem,  since 
the  orbits  are  reentrant  curves.  The  second  is  proved  by 
equation  (130).     The  third  we  will  now  prove. 

5.  Required  the  relation  between  the  time  of  a  complete 
circuit  of  a  particle  in  a7i  ellipse,  and  the  major  axis  of  the 
orbit. 

Let  the  initial  point  be  at  the  extremity  of  the  major  axis 
near  the  pole,  then  6^  in  equation  (d)  will  be  zero,  and  we 
have 

C^  =  firo{l-he); 


190  CENTRAL  FORCES.  [117 1 

but  from  the  ellipse, 

Tq  =z  a  —  ae  =^  a  ().  —  e)  \ 

(a) 


Equation 

(130) 

.-.  C  =  Vfia  (1  -  ^). 
gives 

^       2  area  of  ellipse 

^  -          b 

27r  a2  ^/  (1  -  ^ 

V   fia  (1  -- 

•^ 

6.  The  orbit  being  an  ellipse,  required  the  law  of  the  force 
The  polar  equation  of  the  ellipse,  the  pole  being  at  the  focus, 

is 

_1  _  1  +  g  cos  (^  --  ^o) 
'^~  r~         a{l-  e")         ' 

which,  differentiated  twice,  gives 

d^u  __       e  cos  {6  —  6^ 
W  "  a  (1-  ^     ' 

and  these,  in  equation  (132),  give, 

hence  the  force  varies  inversely  as  the  square  of  the  distance, 

7.  Find  the  law  of  force  by  which  the  particle  may  describe 
a  circle,  the  centre  of  the  force  being  in  the  circumference  of 
the  circle.    (Tait  and  Steele,  Dynamics  of  a  Particle^ 

8.  If  the  force  varies  directly  as  the  distance,  and  is  attrac- 
tive, determine  the  orbit. 

(This  is  the  law  of  molecular  action,  and  analysis  shows  that  the  orbit  is  an 
ellipse.  The  problem  is  of  great  importance  in  Physics,  especially  in  Optica 
and  Acoustics.)   j:  ^,'2-^ 


CHAPTER   IX. 


OONSTEAINED   MOTION    OF   A   PARTICLE. 


118.  If  a  body  is  compelled  to  move  along  a  given  fixed 
curve  or  surface,  it  is  said  to  be  constrained.  The  given  curve 
or  surface  will  be  subjected  to  a  certain  pressure  which  will  be 
normal  to  it. 

If  instead  of  the  curve  or  surface,  a  force  be  substituted  for 
the  pressure  which  will  be  continually  normal  to  the  surface, 
and  whose  intensity  will  be  exactly  equal  and  opposite  to  the 
pressure  on  the  curve,  the  particle  will  describe  the  same  path 
as  that  of  the  curve,  and  the  problem  may  be  treated  as  if  the 
particle  were  free  to  move  under  the  action  of  this  system  of 
forces. 

Let  iV=  the  normal  pressure  on  the  surface,  and 

L  —f{x^  y,  z)  =  0,  be  the  equation  of  the  surface  ; 
Ox,  Oy,  dg  the  angles  between  iVand  the  respective  coor- 
dinate axes. 

Then 


X  +  Nao^dx—  m 


0 


(143) 


y  +  iV^cos  0y  —  7n-^^  =  0\ 

Z  4-  iVcos  Oz  —  m  -r^  =  0  \ 
in  which  the  third  terms  are  the  measures  of  the  resultants  of 


the  axial  components  of  the  applied  forces, 
the  further  discussion  to  forces  in  a  plane. 
plane  of  the  forces,  then  we  have 


m 


m 


^^  d8  ' 


^Y-\-N 


dx 
ds' 


We  will  confine 
Take  cry  m  the 


(144) 


192  CONSTBAINED  MOTION  OF  \n9.l 

Eliminating  JV,  we  find 

'''  \  -lA-  +  -^  \  =  Xdx-\-  Tdy, 


or 


in 


<sy-<S)1-^^^-^^^^^ 


and  integrating,  making  Vq  the  initial  velocity  along  the  path, 
and  V  the  velocity  at  any  other  point,  we  have 

im  (^  -  V)  =AXdx  +  Tdy) ;  (145) 

hence,  the  living  force  gained  or  lost  in  jpassing  from  ons 
point  to  another  is  equal  to  the  work  done  hy  the  impressed 
forces.  If  the  forces  X  and  Y  are  functions  of  the  coordinates 
X  and  y,  and  the  terms  within  the  parenthesis  are  integrable,  the 
result  may  be  expressed  in  terms  of  constants  and  functions  of 
the  coordinates  of  the  initial  and  terminal  points,  and  may  be 
written  ^m{^  —  v^)  =  C(f>  (aJo,2/o)  —  c  0  {x,y) ; 

hence,  for  such  a  system,  the  velocity  will  he  independent  of  the 
path  described^  and  will  he  dependent  only  upon  the  coordinates 
of  the  points ,'  also,  the  velocity  will  he  independent  of  the  nor- 
mal pressure. 

119.  To  find  the  normal  pressure 
multiply  the  first  of  equations  (144)  by  dy,  the  second  by  dx^ 
subtract,  observing  that  do?  -\-  dy^  =  d^,  and  we  find 


dy  d^x  )  _ 
~  ds  df^~ 


\  dx  d?y      dy  d^x  f_  ^dx  _   -^^dy      ^ 
'^]d^lii'~'dsdtH~      d^  d^^      ' 


or 


d^  \  dxd^y  —  dyd^x  \       ^  dx       ^dy  ^    ^.^ 

in  which  p  is  the  radius  of  curvature  at  the  point.  The  first 
and  second  terms  of  the  second  member  are  the  normal  com- 
ponents of  the  impressed  forces.     The  total  normal  pressure 


[120, 121. J  A  PARTICLE.  193 

will,  therefore,  he  that  due  to  the  impressed  forces  plus  that 
due  to  the  force  necessary  to  deflect  the  body  from  the  tangent. 
Tlie  last  term  is  called  the  centrifugal  force,  as  stated  in  Article 
117.  If  the  body  moves  on  the  convex  side  of  the  curve,  the 
last  term  should  be  subtracted  from  the  others  ;  hence  it  might 

be  written  ±  m  — ;  in  which  +  belongs  to  movement  on  the 

concave  arc,  and  —  on  the  convex. 

120.   To  fl/nd  the  time  of  movement^ 
from  equation  (145),  we  have 


-J 

J  S^ 


Vm  ds 


121,  To  find  where  the  particle  wiM  leave  the  constrain- 
ing curve. 

At  that  point  Ar=  0,  which  gives 

which,  combined  with  the  equation  of  the  curve,  makes  known 
the  point. 

If  a  body  is  subjected  only  to  the  force  of  gravity,  we  have 
X=  0  in  all  the  preceding  equations. 


Examples. 

1.  A  hody  slides  down  a  smooth  inclined  plane  under  the 
force  of  gravity  ;  required  the  formulas  for  the  motion. 

Take  the  origin  at  the  upper  end  and  let  the  equation  of  the 
plane  be 

13 


(/ 


194  EXAMPLES.  1121.1 

y  being  positive  downward.     Then  we  have 

JT  =  0,         J^  =  rag^        dy  =  adx,        Vq  =  0. 

and  equation  (145)  becomes 

"ir^  ■=  2fgadx  =  2gax  =  2gy  ;  (a) 

hence,  the  velocity  is  the  same  as  if  it  fell  vertically  through 
the  same  height. 

To  find  the  time,  equation  (147)  gives 

that  is,  if  the  altitude  of  ih^  j^lane  (y)  is  constant  the  ti/m^ 
varies  directly  as  the  length,  s. 

We  may  also  find  y)- -  ^  '^  ^ 

s  =  t  Vigy  =  igf  sin  a.  (c) 

2.  Prove  that  the  times  of  descent  down  all  chords  of  a  ver- 
tical circle  which  pass  through  either  extremity  of  a  vertical 
diameter  are  the  same. 

3.  Find  the  straight  line  from  a  given  point  to  a  given  in- 
clined plane,  down  which  a  body  will  descend  in  the  least  time. 

-  4.  The  time  of  descent  down  an  inclined  plane  is  twice  that 
down  its  height ;  required  the  inclination  of  the  plane  to  the 
horizon. 

5.  At  the  instant  a  body  begins  to  descend  an  inclined  plane, 
another  body  is  projected  up  it  with  a  velocity  equal  to  the 
velocity  which  the  first  body  will  have  when  it  reaches  the  foot 
of  the  plane ;  required  the  point  where  they  will  meet. 

6.  Two  bodies  slide  down  two  inclined  lines  from  two  given 
points  in  the  same  vertical  line  to  any  point  in  a  curve  in  tlie 
same  time,  the  lines  all  being  in  one  vertical  plane ;  required 
the  equation  of  the  curve. 

7.  A  given  weight,  P,  draws  another  weight,  W,  up  an  in- 
clined plane,  by  means  of  a  cord  parallel  to  the  plane ;  through 
what  distance  must  P  act  so  that  the  weight,  W,  will  move  8 
feet  after  P  is  separated  from  it. 


ritt.i 


SIMPLE  PENDULUM. 


195 


8.  Required  a  curve  srcch  that  if  it  revolve  with  a  uniform 
<jmgular  velocity  about  a  vertical  diaweter^ 
and  a  sm/joth  ring  of  infinitesimal  diameter 
he  placed  upon  it  at  any  pointy  it  will  not 
aUde  on  the  curve. 

Let  ft)  be  the  angular  velocity,  then   we 
have 

Y  =  —  mg^       X  —  m(M?x^  ^    t;  =  0, 

and  equation  (145)  becomes  ;. 

ft)V  -  Sp'y  +  6^  =  0, 

which  is  the  equation  of  the  common  parabola. 
If  the  origin  be  taken  at  B^  O  will  be  zero. 

(Another  Solution. — Let  NR  be  a  normal  to  the  curve,  MR  —  the  cen- 
trifugal force,  NM  =  the  force  of  gravity ;  but  the  latter  is  constant,  hence 
iOf,  the  subnormal,  is  constant,  which  is  a  property  of  the  common  parabola.) 

9.  Find  the  normal  pressure  against  the  curve  in  the  pre- 
ceding problem. 

10.  The  Pendulum. — Find  the  time 
of  oscillation  of  the  simple  pendulum. 
This  is  equivalent  to  finding  the  time 
of  descent  of  a  particle  down  a  smooth 
arc  of  a  vertical  circle. 

Take  the  origin   of  coordinates  at  A, 
the  lowest   point.     Let  the  particle  start 
at  D,  at  a  height  AC  =  A;  when  it  has  arrived  at  P,  it  will 
have  fallen  through  a  height  CB  =  h  —  y^  and 
equation  {a)  on  the  preceding  page,  will  have  a  velocity 


,  according  to 


V  =   V2g{h-y)  = 

The  equation  of  the  arc  is 

a^  =  2ry  -  j^; 


ds 

di 


(a) 


hence 


196  EXAMPLES  OP  riM. 

But 


ds  = 


_        rdy 


Considering  this  as  negative,  since  for  tlie  descent  the  arc  is 
a  decreasing  function  of  the  time,  we  have  from  {a) 

T        r^  dy 


This  may  be  put  in  a  form  for  integration  by  Elliptic  Func- 
tions ;  but  by  developing  it  into  a  series,  each  term  may  be 
easily  integrated.     In  this  way  we  find 

■'=W;-ii.(*)t.K0^-(w)'(£)v.-b 

by  means  of  which  the  time  may  be  approximated  to,  with  any 
degree  of  accuracy.  J  c^if 

When  the  arc  is  very  small,  all  the  terms  containing  ^  will 

be   small,  and  by   neglecting  them,  we  have   for  a   complete 
oscillation  (letting  I  be  the  length  of  the  pendulum), 

T=2^  =  7rJl;  {h) 

^    9 

that  isj/br  very  small  arcs  the  oscillations  may  he  regarded  as 
isochronal,  or  performed  in  the  same  time. 

For  the  same  place  the  times  of  vibration  are  directly  as  the 
square  roots  of  the  lengths  of  the  jpendvlums. 

For  any  pendulum  the  times  of  vibration  vary  inversely  as 
the  square  roots  of  the  force  of  gravity  at  different  jplaces. 

If  t  is  constant 

l^  g. 

11.  What  is  the  length  of  a  pendulum  which  will  vibrate 
three  times  in  a  second  ? 

12.  Prove  that  the  lengths  of  pendulums  vibrating  during  the 


[121.]  CONSTRAINED   MOTION  197 

same  time  at  the  same  place,  are  inversely  as  the  squares  of 
the  number  of  vibrations. 

13.  Find  the  time  of  descent  of  a  particle  down  the  arc  of  a 
cycloid. 

The  differential  equation  of  the  curve  referred  to  one  end 
as  an  origin,  x  being  horizontal  and  y  vertical  {r  being  the 
radius  of  the  generating  circle),  is 

dx  =  -- — ' ^  dy. 


v/j 


Ans. 

9 

The  time  will  be  the  same  from  whatever  point  of  the  curve 
the  motion  begins,  and  hence,  it  is  called  tautodtronaL  ?  r;^*b 

14.  In  the  simple  pendulum,  find  the  point  where  the  tension 
of  the  string  equals  the  weight  of  the  particle. 

Ih.  A  jpartiele  is  placed  in  a  smooth  tube  which  revolves 
horizontally  about  an  axis  through  one  end  of  it  /  required 
the  equation  qfthe^urve  traced  hy  the  jparticle. 

The  only  force  to  impel  the  particle  along  the  tube  is  the 
centrifugal  force  due  to  rotation. 

Letting  r  =  the  radius  vector  of  the  curve ; 
7*0  =  the  initial  radius  vector ; 
o)  =  the  uniform  angular  velocity ; 
we  have 

which,  integrated,  gives 

r  =  ir,  (e-  +  e—'), 

hence,  the  relation  between  the  radius  vector  and  the  arc  de- 
scribed by  the  extremity  of  the  initial  radius  vector,  is  the  same 
as  between  the  coordinates  of  a  catenary.  (See  equation  (/&), 
p.  134.) 

16.  To  find  a  curve  joining  two  points  down  whicili  a  par- 
ticle will  slide  by  the  force  of  gravity  in  the  shortest  time. 
The  curve  is  a  cycloid.     This  problem  is  celebrated  in  the 


198  CENTRIFUGAL  FORCE  ri?2,  133.] 

history  of  Dynamics.  The  solution  properly  belongs  to  the 
Calculus  of  Variations,  although  solutions  may  be  obtained  by 
more  elementary  mathematics.  Such  curves  are  called  Bra- 
chistochrones. 


PROBLEMS   PERTAINING    TO    THE   EARTH. 

122.  To  find  the  value  of  g. 

We  have,  from  example  10  of  the  preceding  Article, 

Making  T  =1  second  and  I  =  39.1390  inches,  the  length 
of  the  pendulum  vibrating  seconds  at  the  Tower  of  London, 
we  have  for  that  place, 

g  =  32.1908  feet. 

The  relation  between  the  force  of  gravity  at  different  places 
on  the  surface  of  the  earth  is  given  in  Article  19. 

The  determination  of  I  depends  upon  the  compound  pendu- 
lum. 

123.  To  find  the  centrifugal  force  at  the  equator. 
We  have,  from  equation  (142),  for  a  unit  of  mass, 

f^ofE^'^B;  {a) 

in  which  i?,  the  equatorial  radius,  is  20,923,161  feet;  jT,  the 
time  of  the  revolution  of  the  earth  on  its  axis,  is  86,164  seconds, 
and  TT  =  3.1415926.     These  values  give 

/=r  0.11126  feet. 

The  force  of  gravity  at  the  equator  has  been  found  to  be 
32.09022  feet  (Article  19) ;  hence,  if  it  were  not  diminished 
by  the  centrifugal  force,  it  would  be 

G  =  32.09022  +  0.11126  ^  32.20148  feet, 
and 

/_    0.11126  _    1  , 

^-32.20148.-289''^^''^^' 


[134.1  ON  THE  EARTH.  199 

hence  the  centrifugal  force  at  the  equator  is  ji^  ^^  the  un- 
diminislied  force  of  gravity. 

Example. 

In  what  time  must  the  earth  revolve  that  the  centrifugal 
force  at  the  equator  may  equal  the  force  of  gravity  ? 

Ans.  ^  of  its  present  time, 

124,  To  find  the  effect  of  the  centrifugal  force  at  different 
latitudes  on  the  earth. 

Let  L  =  POQ  =  the  latitude  of  the  point  P  ; 

B  =z  OQ  =  OP  —  the  radius  of  the  earth ; 

then  will  the  radius  of  the  parallel  of  lati-  

tude  pp'  be  ^.^,<:::rz>^p>^.^ 

7j?i  =  E  cos  L,  fei±^^<f 

The   centrifugal   force   will   be   in   the  \^  ^ 

plane  of  motion  and  may  be  represented  pio.  ii4, 

by  the  line  Pr^  or 

Pr  =  fi  =  (o^Ei  =  (o^EcosZ; 

therefore,  the  centrifugal  force  varies  directly  as  the  cosine  of 
the  latitude.  But  the  force  of  gravity  is  in  the  direction  PO. 
llesolving  Pr  parallel  and  perpendicular  to  PO,  we  have 

Pj>  =  Q)^P  cos^  Z  =  -2  J^  G  cos^  Z ; 

Pq  =  arP  cos  Z  sin  Z  =  ^fy  G  sin  2Z  ; 

the  former  of  which  diminishes  directly  the  force  of  gravity, 
and  the  latter  tends  to  move  the  matter  in  the  parallel  of  lati- 
tude PP\  toward  the  equator.  Such  a  movement  has  taken 
place,  and  as  a  result  the  earth  is  an  oblate  spheroid.  In  the 
present  form  of  the  earth  the  action-line  of  the  force  of  gravity 
is  normal  to  the  surface  (or  it  would  be  if  the  earth  were  homo- 
geneous), and  hence,  does  not  pass  through  the  centre  O,  except 
on  the  equator  and  at  the  poles.  The  preceding  formulas 
would  be  true  for  a  rigid  homogeneous  sphere,  but  are  only 
approximations  in  the  case  of  the  earth. 


^ 


CHAPTEK  X. 


FOKOES   IN  A  PLANE  PEODUOING  EOTATION. 


125,  Angular  motion  of  a  particle  about  a  fixed  axis. 

Let  the  body  G^  on  the  horizontal  arm  AB^  revolve  about  .the 
vertical  axis  ED.  Consider  the  body 
reduced  to  the  centre  of  the  mass, 
and  the  force  F^  applied  at  the  centre 
and  acting  continually  tangent  to  the 
path  described  by  the  particle.  This 
may  be  done  as  shown  in  Fig.  121. 
In  this  case  the  force  will  be  measured 
in  the  same  way  as  if  the  path  were 
rectilinear,  for  the  force  is  applied 
Hence,  according  to  equation  (21), 


along  the  path. 


in  which  8  is  the  arc  of  the  circle. 

If  B  be  the  angle  swept  over  by  the  radius,  and  r^  the  radius ; 
then 

5  =  n  ^, 
ds  =^  r^  dO, 

which,  substituted  in  the  equation  above,  gives 


F^  =  Tnvx 


(Pd_ 
df' 


d?e 

•**  de 


(150) 


If  a  force,  F,  be  applied  to  the  arm  AB,  at  a  distance  a  from 


[126, 127.]  ANGULAE  ACCELERATION.  201 

the  axis  ED^  producing  the  same  movement  of  the  mass  (7,  we 
have,  from  the  equality  of  moments,  Article  65, 

and  the  value  of  F^  deduced  from  this  equation  substituted  in 
the  preceding  one,  gives 

that  is,  the  angular  acceleration  produced  by  a  force,  F,  on  a 
jpartide,  m,  equals  the  moment  of  the  force  divided  hy  the  mo- 
ment of  inertia  of  the  mass, 

(For  moments  of  inertia,  see  Chapter  YII.) 

We  observe  that,  when  the  force  is  applied  directly  to  the 
particle,  it  produces  no  strain  upon  the  axis,  but  that  it  does 
when  applied  to  other  points  of  the  arm.  In  both  cases  there 
will  be  a  strain  of 

due  to  centrifugal  force,  a  being  the  angular  velocity. 


126,  Angui^ae  motion  of  a 
FINITE  MASS.  Let  a  body,  AB, 
turn  about  a  vertical  axis  at  A, 
under  the  action  of  constant 
forces,  F,  acting  horizontally ; 
mi,  7/12,  6tc.,  masses  of  the  ele- 
ments of  the  body  at  the  re-  ^V.,  ,  ...-'''' 
spective  distances  ri,r^  etc.,  no.  iw. 
from  the  axis  A  ;  and  consid- 
ering equation  (151)  as  typical,  we  have 

(Pd XFa ^Fa 

dt^  ~~  mi/'i2+  m^r^-\-  m^ri  +  etc.,  "~  Xmr^ 

moment  of  the  forces  ,^  ^^. 

m^me?it  of  iiiertia    '  ^ 

127.  Eneegt  of  a  kotating  mass.    Multiply  both  members 


202  FORMULAS  FOR  [128, 129.1 

of  the  preceding  equation  by  dd^  integrate  and  reduce,  and  we 
find 


iSm/^g)'    =.fF,rdd',  (153) 


in  which  rdO  is  an  element  of  the  space  passed  over  by  F^  and 
F.rdO  is  an  element  of  work  done  by  F\  hence  the  second 
member  represents  the  total  work  done  by  i^upon  the  body,  all 
of  which  is  stored  in  it.  Therefore,  the  energy  of  a  body  rotat- 
ing about  an  axis  equals  the  moinent  of  inertia  of  the  Tnass 
multijplied  by  one-half  the  square  of  the  angular  velocity. 

If  tlie  body  has  a  motion  of  translation  and  of  rotation  at  the 
same  time,  the  total  energy  will  be  the  sum  due  to  both  motions ; 
for  it  is  evident  that  while  a  body  is  rotating  a  force  may  be 
applied  to  move  it  forward  in  space,  Article  3S,  and  that  the 
work  done  bj^  this  force  will  be  independent  of  the  rotation. 
\iv  —  the  velocity  of  translation  of  the  axis  about  which  the 
body  rotates ; 
ft)  =  the  angular  velocity ;  and 
I^  =  the  moment  of  inertia  of  the  mass; 
then  the  total  work  stored  in  the  body  will  be 

kMv'  +  i/^a>2  (154) 

128,  ^N    IMPULSE.      Multiply   both   members   of    equation 
(152)  by  dt,  integrate,  and  we  find 

de^  _      _  tafFdt 
dt  ~~      ~~    Xmn^ 

But  according  to  Article  27,  fFdt  is  the  measure  of  an  im- 
pulse and  is  represented  by  Q^  hence 

_  moment  of  the  impulse 
""      Wjoment  of  inertia 

129.  The  time  required  to  pass  over  n  circumferences  will 
be,  m  the  case  of  an  impulse, 


[130,  131.]  ANGULAR  ROTATION. 

__  2n7r  _  ^iirT 
~~     CO    ~~     Qa 

In  the  case  of  accelerated  forces  the  time  will  be  found  by 
integrating  equation  (153). 

130.  Simultaneous  movement  of  rotation  and  of  transla- 
tion. If  the  body  be  unconstrained,  the  motion  of  transla- 
tion of  the  body  will  be  that  of  the  centre  of  the  mass.  If  the 
axis  of  rotation  is  rigid,  it  may  be  located  any  wliere  in  the  body, 
or  even  without  the  body  by  considering  it  as  rigidly  connected 
with  the  body,  in  which  case  the  motion  of  translation  will  be 
that  of  some  point  of  the  axis.  In  either  case  the  motion  of 
translation  may  be  considered  as  resulting  from  a  force  acting 
directly  upon  the  axis  of  i-otation,  and  the  rotation,  by  a  force 
acting  at  some  other  point.  The  two  motions  may  then  be 
considered  as  existing  independently  of  each  other. 

131.  Formulas  for  the  movement  of  a  body  involving  both 
translation  and  rotation.  The  general  equations  for  this  case 
are  (164)  and  (165)  in  the  next  chapter.  In  this  Artic^le  let  the 
rotation  be  about  an  axis  parallel  to 2,  and  the  centre  of  the  mass 
move  in  the  plane  xy,  Kesolve  the  forces  into  couples  and  forces 
applied  at  the  origin  of  coordinates,  as  in  Article  83 ;  then  will  the 
third  of  equations  (86)  be  the  impressed  forces  which  produce 
rotation.  Let  E  be  the  resultant  of  the  forces  at  the  origin  at 
any  instant,  (see  Article  84),  and  s  be  measured  along  the  path 
described  by  the  point  of  intersection  of  the  axis  of  rotation 
with  the  plane  xy.    Then  will  the  principles  of  Article  21,  give 


dr 


J(X,-ra,)-z(«.,5-^g)  =  0; 


(156) 


The  expression  X  (Xy  —  Tx)  is  the  sum  of  the  moments  of 
the  impressed  forces  =  XFa  (Article  60).  Transforming  the 
second  term  of  the  last  equation  into  polar  coordinates  having 
the  same  origin,  we  have 

cPx  rPy\  _  ^        d^e 


X[m/-£^-m,,^^=Xm 


d^' 


204 


REDUCED  MASS. 


[132.1 


hence,  the  equations  become 


7)1 


df 
dW 

df 


=  F; 


ZFa 


(157) 


132,  Keduced  mass.  A  given  mass  may  be  concentrated 
at  such  a  point,  or  in  a  thin  annulus,  that  the  force  or  impulse 
will  have  the  same  effect  upon  it  as  if  it  were  distributed.  To 
accomplish  this  it  is  only  necessary  that  Ximr^  in  the  second  of 
(157)  should  have  an  equivalent  value.  Let  if  be  the  mass  of  the 
body,  Jc  the  distance  from  the  axis  to  the  required  point,  then 

Xmr^  =  JIf F, 
in  which  Tc  is  the  radius  of  gyration,  as  defined  in  Article  107. 
.But  any  other  point  may  be  assumed,  and  a  mass  determined 
such  that  the  effect  shall  be  the  same.  Let  k  be  the  distance  to 
that  point  (or  radius  of  the  annulus),  and  M^  the  required  mass 
then  we  have 


which  is  called  the  reduced  mass. 


/P 


(158) 


^ 


^ 


^^ 


Examples. 

1.  A  slender  tar  AB,  falls  through  a  height  h,  retaining 
its  horizontal  position  until  one  end  strikes  a  fixed  obstacle  C ; 
required  the  angular  velocity  of  the  jpiece  and  the  linear  velo- 
city oftlie  centre  immediately  after  the  impulse. 

Let  M  be  the  mass  of  the  bar,  I  its 
length,  V  the  velocity  of  the  centre 
at  ^he  instant  of  impact,  and  v^  the 
velocity  of  the  centre  immediately 
after  impact.  Consider  the  bodies  as 
perfectly  non-elastic;  then  will  the 
effect  of  the  impact  be  simply  to  sud- 
denly arrest  the  end  A. 

The  bar  will  rotate  about  a  horizontal  axis  through  the  centre, 
as  shown  by  Article  38 ;  and,  as  shown  by  Articles  27  and  38,  the 


Fig.  117. 


[132.]  EXAMPLES.  205 

impulse  will  be  Q  =  M  (v  —  Vi);  that  is,  it  is  the  change  of 
velocity  at  the  centre  multiplied  hy  the  mass.  Tlie  impact  will 
entirely  arrest  the  motion  of  the  end,  A,  at  the  instant  of  tho 
impact,  and  hence  at  that  instant  the  angular  velocity  of  A  in 
reference  to  G  will  be  the  same  as  G  in  reference  to  A, 
Equation  (155)  gives 

_  TTioment  of  impulse 


moment  of  inertia 

M  {v  -  V,)  U 

^.,MJ? 

6^-^^ 

=  6       ^     . 

But  at  the  instant  of  the  impact 

solving  these  give 

We  now  readily  find 

Q  =  \Mv, 
To  find  the  velocity  of  any  point  in  a  vertical  direction  at 
the  instant  of  the  impact,  we  observe  that  it  may  be  considered 
as  composed  of  two  parts ;  a  linear  velocity  v^  downward,  and  a 
right-handed  rotation.  The  actual  velocity  at  A  due  to  rota- 
tion will  be 

which  will  be  upward,  and  the  linear  velocity  downward  will  be 
"^i  =  f^>  hence  the  result  will  be  no  velocity.  Similarly,  the 
velocity  at  B  will  be  Jt;  +  ft;  =  |v.  Also,  for  any  point  dis- 
tant X  from  G^  we  have  at  the  left  of  G 

J?>  —  wa?  =   f^'(l  —  2  ^J; 

and  to  the  right  of  G  we  have 

1.(1  +  2^). 
When  the  bar  comes  into  a  vertical  position,  we  easily  find 


206 


ANGULAR  MOVEMENT. 


[132.] 


that  A  has  passed  below  a  horizontal  through  0.  Every  point, 
therefore,  has  a  progressive  velocity,  except  the  point  A,  at 
the  instant  of  impact. 

After  the  impact  the  centre  will  move  in  the  same  vertical 
and  with  an  accelerated  velocity,  while  the  angular  velocity 
will  remain  constant. 

2.  Suppose  that  impact  takes  place  at  one-quarter  the  length 
from  A,  required  the  angular  velocity. 

3.  At  what  point  must  the  impulse  be  made  so  that  the  velocity 
of  the  extremity  JS  will  be  doubled  at  the  instant  of  impact  ? 

4.  An  inextensible  string  is  wound  around  a 
cylinder,  and  has  its  free  end  attached  to  a 
fixed  point.  The  cylinder  falls  through  a  cer- 
tain height  (not  exceeding  the  length  of  the  free 
part  of  the  string),  and  at  the  instant  of  the  im- 
pact the  cord  is  vertical  and  tangent  to  the 
cylinder ;  all  the  forces  being  in  a  plane ;  re- 
quired the  angular  velocity  produced  by  the 
impulse,  and  the  momentum. 


Pio.  118. 


A7is.%-;     Q 


iMv, 


6.  In  the  preceding  problem,  let  the  body  be  a  homogeneous 
sphere,  the  string  being  wound  around  the  arc  of  a  great 
circle. 


A  Q-M  v. 


W^ 


W^^m\ 


B 


6.  A  homogeneous  _pris7natiG 
bar  AJB,  in  a  horizontal  posi- 
tion C071  strained  to  revolve  about 
a  vertical  fixed  axis  A,  receives 
a  direct  impulse  from  a  sphere 
whose  momentum  is  Mv ;  re- 
quired  the  anguh^r  velocity  of 
the  bar. 
The  momentum  imparted  to  the  bar  will  depend  upon  the 

elasticities  of  the  two  bodies.     Consider  them  perfectly  elastic. 

The  effect  of  the  impulse  will  be  the  same  as  if  the  mass  of  the 

bar  were  concentrated  at  the  extremity  of  the  radius  of  gyration ; 

hence  an  equivalent  mass  at  the  point  G  may  be  determined. 


Fig.  119. 


[132.]  EXAMPLES.  307 

Let  Mx  =  the  mass  of  AB ; 
M^  =  the  reduced  mass ; 

V2  =  the  velocity  of  the  reduced  mass  after  impact ; 
a  =  AC. 
Then,  by  equation  (158),  the  mass  of  the  bar  reduced  to  the 
point  6^,  will  be 

a*' 
JbSy  equation  (40)  the  velocity  of  M^  after  impact  will  be 

hence,  the  momentum  imparted  will  be 

2JOfi/5r» 

and  the  moment  will  be 

According  to  equation  (155),  we  lia^e 

_  moment  of  the  impulse 
~~      moment  ofine^'tia 


Tliis  result  is  the  same  as  that  found  by  dividing  equation 
{a)  by  a,  as  it  should  be. 

7.  Suppose^  in  the  preceding  prohlem,^  that  there  is  no  fixed 
axis,  hut  that  the  hody  is  free  to  translate  /  find  where  the  im- 
pact must  he  made  that  the  initial  velocity  at  the  end  A  shall 
he  zero. 


(a\ 


20S  ANGULAR  MOVEMENT.  [133.1 

Let  Mv  be  the  impulse  imparted  to  the  body; 
Mk^  =  the  principal  moment  of  inertia ; 

h  =  the  distance  from  the  centre  of  the  bar  to  the 
required  point ; 
then 

_  moment  of  the  imptilse 
inoment  of  inertia 

_  Mvh  _  yA 

and  the  movement  at  A  in  the  circular  arc  will  be 

.,  nlh 

*^  =  wr 

and  the  initial  linear  movement  will  be 

vlh 

which,  by  the  conditions  of  the  problem,  will  be  zero ;  hence 

vlh 

or, 

h  =  ^J^.  (6) 

The  distance  from  A  will  be 

If  the  bar  be  of  infinitesimal  section 

The  result  is  independent  of  'the  magnitude  of  the  impulse. 
From  (5)  we  have 

hence,  h  and  ^  are  convertible,  and  we  infer  that  if  the  im« 
pulse  be  applied  at  A  the  point  of  no  initial  motion  will  be  at 


[183.]  AXIS  OF  SPONTANEOUS  ROTATION.  209 

the  point  given  by  equation  (5),  where  the  impact  was  previ- 
ously applied. 

8.  In  the  ^preceding prohlem  find  where  the  impulse  inust  he 
applied  so  that  the  point  of  no  initial  velocity  shall  he  at  a 
distance  h'  from  the  centre. 

The  initial  linear  velocity  due  to  the  rotary  movement  found 
from  {a)  of  the  preceding  example,  will  be 

h'(0  =  j-^  hh\ 

and  the  initial  movement  of  the  required  point  being  zero,  we 
have 

.-.  A  =  f.  (159) 

If  the  point  of  impact  be  at  J,  the  point  «,  where  the  initial 

movement  is  zero,  will  be  on  the  other       .                 v  n 

side  of  the  centre  of  the  body.     Let  B      '    ^            '  '6       ' 
be  the  centre,  then                                                     ^'°-  ^^• 

h  =  hB,        h'  =  aBy 
and  from  (159),  we  have 

hh'  =  hB.aB  =  ki^ ;  (160) 

and  as  ^  is  a  constant,  the  points  a  andh  are  convertihle. 

AXIS    OF    SPONTANEOUS    ROTATION. 

133.  In  the  preceding  problem  the  initial  motion  would 
liave  been  precisely  the  same  if  there  had  been  a  fixed  axis 
through  a  perpendicular  to  the  plane  of  motion,  and  hence  the 
initial  motion  may  be  considered  as  a  rotation  about  that  axis. 
If  a  fijJcd  axis  were  there  it  evidently  would  not  receive  any 
shock  from  the  impulse. 

The  axis  ahout  which  a  quiescent  hody  tends  to  turn  at  the 
instant  that  it  receives  an  impidse  is  called  the  axis  of  spon- 
taneous  rotation, 
14 


210  INSTANTANEOUS  AXIS.  ri34,135.j 

OKNTRE    OF   PERCUSSION. 

134.  When  there  is  a  fixed  axis  and  the  body  is  so  struck 
that  there  is  no  impulse  on  the  axis,  any  jpoint  in  the  aGtign- 
line  of  the  force  is  called  the  centre  of  percussion.  Thus  in 
Fig.  120,  if  a  is  the  fixed  axis,  h  will  be  the  centre  of  percus- 
sion. It  is  also  evident  that,  if  J  be  a  fixed  object,  and  it  be 
struck  by  the  body  AG,  rotating  about  a,  the  axis  will  not 
veceive  an  impulse. 

AXIS    OF   INSTANTANEOUS    ROTATION. 

135.  An  axis  through  the  centre  of  the  mass,  parallel  to  the 
axis  of  spontaneous  rotation,  is  called  the  axis  of  instantaneotos 
rotation.    A  free  body  rotates  about  this  axis. 

In  regard  to  the  spontaneous  axis,  we  consider  th^it  as  fixed 
in  space  for  the  instant ;  but  at  the  same  time  the  body  really 
rotates  about  the  instantaneous  axis  which  moves  in  space  with 
the  body. 

EXAMPLES  UNDER  THE  PRECEDING  EQUATIONS  CONTINUED. 

9.  A  horizontal  uniform  disc  is  free  to  revolve  about  a  ver- 
tical axis  through  its  centre.  A  man  walks  around  on  the 
outer  edge  •  required  the  angular  distance  passed  over  hy  the 
man  and  disc  when  he  has  walked  once  around  the  circumjer- 
ence. 

Let  W  =  the  weight  of  the  man  ; 
w  =  the  weight  of  the  disc ; 
r  =  the  radius  of  the  disc ; 
G)i  =  the  angular  velocity  of  the  man  in  reference  to  a 

fixed  line  ; 
o)  =  the  angular  velocity  of  the  disc  in  reference  to  the 

same  fixed  line ; 
Q  =  the  force  exerted  by  the  man  against  the  disc ; 

The  result  will  be  the  same  whether  the  effort  be  exerted 
suddenly,  or  with  a  uniform  acceleration,  or  irregularly.  We 
will,  therefore,  treat  it  as  if  it  were  an  impulse.  Tlie  weights 
are  here  used  instead  of  the  masses,  for  they  are  directly  pro- 


(135.]  EXAMPLES.  311 

portional  to  each  other,  and  it  is  more  natural  to  speak  of  the 
weight  of  a  man  than  the  mass  of  a  man. 
We  have 

__  moment  of  the  iiriptdse 
""      moment  of  inertia 

^  Qr 

wk} 

_  Wvr 

2W 
=  —  ©1. 

w 

If,  in  a  unit  of  time  the  man  arrives  at  the  initial  point  of 
the  disc,  we  have 

0)  -f  fi>i  =  27r ; 
which,  combined  with  the  preceding  equation,  gives 

_      ^wir 

If  IF  =  to  J  we  have 

for  the  angular  space  passed  over  by  the  man,  and 

0)   =   f  TT, 

for  the  distance  passed  over  by  the  disc. 

10.  In  Fig.  115  let  the  force  F  he  constant;  required  the 
number  of  cornpleie  turns  which  the  body  C  will  make  about 
the  axis  DE  in  the  time  t. 

Let  r  =  the  radius  of  the  circle  passed  over  by  F\ 

Tx  =  the  distance  of  the  centre  of  the  body  from  the  axis 

of  revolution ; 
h^  =  the  principal  radius  of  gyration  of  the  body  in  refer- 
ence to  a  moment  axis  parallel  to  DE\ 
k  =  the  radius  of  gyration  of  the  body  in  reference  to 
the  axis  DF; 
then,  according  to  equation  (123), 

^  =  r,'  +  k,'; 


212 


ROTATION  OP 


ridc.i 


and,  according  to  equation  (152), 

d^Q  _  Tnoment  of  forces 
df       moment  of  inertia 

_    Ft 

Multiply  by  dt  and  integrate,  and  we  have 

de  _  Fr 

dt  ~  Ml^ 

the  constant  being  zero,  for  the  initial  quantities  are  zero. 
Multiplying  again  by  dt^  we  iind 

Fr 


e 


'ZMk' 


^; 


which  is  the  angular  space  passed  over  in  time  t ;  and  the  num- 
ber of  complete  rotations  will  be 

e_  _     Frf 

11.  If  the  body  were  a  sphere  2  feet  in  diameter,  weighing 
100  pounds,  the  centre  of  which  was  5  feet  from  the  axis ;  F,  a 
force  of  25  pounds,  acting  at  the  end  of  a  lever  8  feet  long ; 
required  the  number  of  turns  which  it  will  make  about  the  axis 
in  5  minutes. 

12.  If  the  data  be  the  same  as  in  the  preceding  example ; 
required  the  time  necessary  to  make  one  complete  turn  about 
the  axis. 

13.  Suppose  that  an  indefinitely  thin  body,  whose  weight  is  Tf, 

rests  upon  the  rim  of  a  horizon- 
tal pulley  which  is  perfectly  free 
to  move.  A  string  is  wound 
around  the  pulley,  and  passes 
over  another  pulley  and  has  a 
weight,  P,  attached  to  its  loWer 
end.  Supposing  that  there  is  no 
resistance  by  the  pulleys  or  the 
string,  required  the  distance 
passed  over  by  P  in  time  t. 


Fio.  121 


[185.1  SOLID  BODIES. 

According  to  equation  (152),  we  have 

d^e  Ft P£ 


213 


df    ~  TTTT^  ~  {W+  F)r' 


from  which  it  may  be  solved. 

(This  is  equivalent  to  applying  the  weight  P  directly  to  the  weight  TT,  oa 
in  Pig.  10,  and  hence  we  have,  according  to  equation  (21), 

W+P  dh 


9 


d^ 


=  P; 


but  referring  it  to  polar  coordinates,  we  have  r 


(P9  _  d-8 

dF~d^ 


,  which  substituted 


reduces  the  equation  directly  to  that  in  the  text. ) 

14-.  A  disc  whose  weight  is  TFis  free 
to  revolve  about  a  horizontal  axis  pass- 
ing through  its  centre  and  perpendicular 
to  its  plane.  A  cord  is  wound  around 
its  circumference  and  lias  a  weight,  P, 
attached  to  its  lower  end  ;  required  the 
distance  through  which  P  will  descend 
in  t  seconds. 

We  have 

(pe_  __         Prg 

df    "  Wk;'  4-  Pr^' 

from  which  6  may  be  found,  and  the  space  will  be  rO. 

(This  may  be  solved  by  equation  (21).     The  mass  of  the  disc  reduced  to  an 


W  k  ' 
equivalent  at  the  circumference  will  be  —   — 

9    T'^ 


and  that  equation  will  become 


1  /  A;  '\  d?% 

—{P-\-  ^—7-  J-jTj  =  P\  which,  by  chang^g  to  polar  coordinates,  may  be 

reduced  to  the  equation  in  the  text.)  • 

15.  If,  in  the  preceding  example,  the  body  were  a  sphere 
revolving  about  a  horizontal  axis,  the  diameter  of  the  spliere 
being  16  inches,  weight  500  pounds,  moved  by  a  weight  of  100 
pounds  descending  vertically,  the  cord  passing  around  a  groove 
in  the  sphere  the  diameter  of  whi(;h  is  one  foot ;  required  the 
number  of  revolutions  of  the  sphere  in  five  seconds. 


214 


ROTATION  OP 


[13.^) 


Fio.  123. 


16.  Two  weights,  P  and  W,  are  suspended 
on  two  pulleys  by  means  of  cords,  as  shown  in 
Fig.  123,  the  pulleys  being  attached  to  the 
same  axis  O.  No  resistance  being  allowed 
for  the  pulleys,  axle,  or  cords ;  required  the 
circumstances  of  motion. 

We  have 


d^O  _  moment  of  the  forces 
df    ~    moments  of  inertia 

P.AG  -W,BG 

~  P{AC)^  +  Wi^BOf^disoACk^  -f  discBCJc^^' 

in  which  disc  A  0,  etc.,  are  used  for  the  weights  of  the  discs. 
Let  the  right-hand  member  be  represented  by  M,  then  we  have 


dd 

dt 

e 


CO 


Mt; 


\M:f. 


17.  In  the  preceding  example  let  the  discs  be  of  uniform 
density,  ^(7=8  inches,  BC=^  ^  inches ;  the  weight  of  -^  6^  =  6 
pounds,  of  CB  =  2  pounds,  of  P  =  25  pounds,  and  of  TT  =  60 
pounds ;  if  they  start  from  rest,  required  the  space  passed  over 
by  P  in  10  seconds,  and  the  tension  of  the  cords. 

18.  A  homogeReoits^  hollow  cylinder  rolls  down  an  inclined 
plane  hy  the  force  of  gravity  ;  required  the  time. 

The  weight  of  the  cylinder 
may   be    resolved    into    two 
components,  one  parallel  to 
the   plane,  which  impels  the 
body  down  it,  the  other  nor- 
mal, which  induces   friction. 
The  friction  acts  parallel  to 
the  plane  and  tends  to  prevent  the  movement  down  it,  and  1a 
assumed  to  be  sufficient  to  prevent  sliding. 
Let   W  =  the  weight  of  the  cylinder ; 

i  =  tlie  inclination  of  the  plane  to  tlie  horizontal ; 
N :=.  TFcos  i  =  the  normal  component; 
m.  —  the  mass  of  a  unit ;  the  altitude  =  1 ; 


Fig.  124. 


fl85.1  SOLID  BODIES.  215 

<f>  =  the  coefficient  of  friction ; 
T  =  (t>N  =  the  tangential  component ; 
T'  =  W  sin  i  =  the  component  of  the  weight  parallel 

to  the  plane ; 
7*1  =  the  internal  radius  of  the  cylinder ; 
r  =  the  external  radius ; 

6  =  the  angular  space  passed  over  by  the  radius  ; 
8  =  AC,  the  space. 
This  is  a  case   of  translation  and  rotation   combined,  and 
agnations  (157)  give 

-5^  =T'  -  T=  Wsmi-  T; 
g  df  ' 

d'^e  _  T.r  _         ^gTr 


and  from  the  problem 

8  =  rd. 
Eliminating  8  and  T  from  these  equations,  we  get 
d^d  _    2^^7'sin  i 
df  ~  37^4^/7* 

Integrating  and  making  the  initial  spaces  zero,  we  have 
^_    gr  Bin  i   ^ 


/3^ 


4-  n* 

-; 8. 

gi"^  sm  % 


If  ri  =  0,  the  cylinder  will  be  solid,  and 


3* 


^  sm  % 

and  hence,  the   time  is  independent  of  the  diameter  of  the 
cylinder. 

If  r,  =  r,  the  cylinder  will  be  a  thin  annulus,  and 


/     4« 


g  sm  % 
lience,   the    time    of    descent   will    be  V'j  times   as  long  as 


216 


THE  COMPOUND 


[185.1 


when  the  cylinder  is  solid ;  the  weight  being  the  same  in  both 
cases. 

If  it  slide. down  a  smooth  plane  of  the  same  slope,  we  have 


^=v^.-- 


/     2^ 
g  sin  % 

which  is  less  than  either  of  the  two  preceding  times. 


THE    PENDULUM. 

X        19.  Let  a  hody  he  suspended  on  a  horizuiv- 
tal  axis  and  moved  by  the  force  of  gravity ; 
required  the  cireumstances  of  motion. 
We  have 

d^d  __  moment  of  forces 
dt^        moment  of  inertia 

_  Wh  sin  0 

Pio.  125  -^^ 

in  which 

h  —  Oa,  the  distance  from  the  axis  of  suspension  to  the 

centre  of  gravity  a  of  the  body ; 
W  =  the  weight  of  the  body  ; 
0  =  bOa;  and  let 

ki  =  the  principal  radius  of  gyration ; 
then  the  preceding  equation  becomes 

d^d  gh        .    ^ 

~dF  =  WVk?''''^' 

This  equation  cannot  be  completely  integrated  in  finite 
terme,  but  by  developing  sin  0  and  neglecting  all  powers  above 
the  first,  we  lind  for  a  complete  oscillation 


TT 


v 


h^  +  h^ 


gh 


(IGJ) 


which  gives  the  time  in  seconds  when  A,  ^  and  g  are  given  in 
feet. 

To  find  the  length  of  a  simple  pendulum  whicli  will  vibrate 


[136,  137.]  PENDULUM.  217 

ill  the  same  time,  we  make  equations  (b),  page  196,  and  (161) 
equal  to  one  another,  and  have 

I  =  ^U^  =  Od.  (162) 


• 


Let  ad  =  /<i,  then 

Z  -  A  =  Ai  =  ^  ; 

.      .-.    h/i,  =  k^  (163) 

186,  Definitions.  A  body  of  any  form  oscillating  about  a 
fixed  axis  is  called  a  compound  j)endulum. 

A  material  particle  suspended  by  a  string  without  weight, 
oscillating  about  a  fixed  axis,  is  called  a  simjile  pendulum. 

The  point  d  is  called  the  centre  of  oscilhition.  It  is  the 
point  at  which,  if  a  particle  be  placed  and  suspended  from  the 
axis  0  by  a  string  without  weight,  it  will  oscillate  in  the  same 
time  as  the  body  Od.  Or,  it  is  the  point  at  which,  if  the  entire 
mass  be  concentrated,  it  will  oscillate  about  the  axis  in  the  same 
time  as  when  it  is  distributed. 

The  point  O^  where  the  axis  pierces  the  plane  xy^  is  called 
the  centre  of  sicsjpension, 

137.  Results.  The  centres  of  oscillation  and  of  percussion 
wincide.     (See  Article  134.) 

According  to  equation  (163),  the  centres  of  oscillation  and  of 
suspension  are  convertible. 

According  to  the  same  equation  the  principal  radius  of  gyra- 
tion is  a  mean  proportional  between  the  distances  of  the  cen- 
tres of  oscillation  and  of  suspension  from  the  centre  of  gravity. 

Equation  (161)  indicates  a  practical  mode  of  determining  the 
principal  radius  of  gyration.  To  find  it,  let  the  body  oscillate, 
and  thus  find  T,  then  attach  a  pair  of  spring  balances  to  the 
lower  end  and  bring  the  body  to  a  horizontal  position,  and  find 
how  much  the  scales  indicate ;  knowing  which,  the  weight  of  the 
body  and  the  distance  between  the  point  of  attachment  and  the 
centre  of  suspension  O^  the  value  of  h  may  easily  be  computed. 
The  value  of  g  being  known,  all  the  quantiti^  in  equation 
(161)  become  known  except  ^i,  which  is  readil^ound  by  a 
solution  of  the  equation. 


218  CAPTAIN  EATER'S  [138.1 

Examples  . 

1,  A  prismatic  bar  oscillates  about  an  axis  passing  through 
one  end,  and  perpendicular  to  its  length ;  required  the  length 
of  an  equivalent  simple  pendulum. 

2.  A  homogeneous  sphere  is  suspended  from 
a  point  by  means  of  a  fine  thread,  find  the  length 
of  a  simple  pendulum  which  will  oscillate  in  the 
same  time. 

138,  Captain  Kater  used  the  principle  of  the 
convertibility  of  the  centres  of  suspension  and 
oscillation  for  determining  the  length  of  a  simple 
Fig.  126.  seconds  pendulum,  and  hence  the  acceleration 

due  to  gravity. — Phil.  Trans. ^  1818. 

Let  a  body,  furnished  with  a  movable  weight,  be  provided 
with  a  point  of  suspension  O  (figure  not  shown),  and  another 
point  on  which  it  may  vibrate,  fixed  as  nearly  as  can  be  esti- 
mated in  the  centre  of  oscillation  (9,  and  in  a  line  with  the 
point  of  suspension  and  the  centre  of  gravity.  The  oscillations 
of  the  body  must  now  be  observed  when  suspended  from  C  and 
also  when  suspended  from  0,  If  the  vibrations  in  each  y)0?>\- 
tion  should  not  be  equal  in  equal  times,  they  may  readily  be 
made  so  by  shifting  the  movable  weight.  When  this  is  done, 
the  distance  between  the  two  points  6^  and  O  is  the  length  of 
the  simple  equivalent  pendulum.  Thus  the  length  L  and  the 
corresponding  time  T  oi  vibration  will  be  found  uninfiuenced 
by  any  irregularity  of  density  or  figure.  In  these  experiments, 
after  numerous  trials  of  spheres,  etc.,  knife  edges  were  pre- 
ferred as  a  means  of  support.  At  the  centres  of  suspension  and 
oscillation  there  were  two  triangular  apertures  to  admit  the 
knife  edges  on  which  the  body  rested  while  making  its  oscil- 
lations. 

Having  thus  the  means  of  measuring  the  length  Z  with 
accuracy,  it  remains  to  determine  the  time  T.  This  is  effected 
by  comparing  the  vibrations  of  the  body  with  those  of  a  clock. 
The  time  of  a  single  vibration  or  of  any  small  arbitrary  number 
of  vibrations  cannot  be  observed  directly,  because  this  would 
require  the  fraction  of  a  second  of  time,  as  shown  by  the  clock, 
to  be  estimated  either  by  the  eye  or  ear.     The  vibrations  of  the 


[138.]  EXPERIMENTS.         »  219 

body  may  be  counted,  and  here  there  is  no  fraction  to  be  esti- 
mated, but  these  vibrations  will  not  probably  fit  in  with  the 
oscillations  of  the  clock  pendulum,  and  the  differences  must  be 
estimated.  This  defect  is  overcome  by  "  the  method  of  coinci- 
dences." Supposing  the  time  of  vibration  of  the  clock  to  be  a 
little  less  than  that  of  the  body,  the  pendulum  of  the  clock  will 
gain  on  the  body,  and  at  length  at  a  certain  vibration  the  two 
will  for  an  instant  coincide.  The  two  pendulums  will  now  be 
seen  to  separate,  and  after  a  time  will  again  approach  each 
other,  when  the  same  phenomenon  will  take  place.  If  the  two 
pendulums  continue  to  vibrate  with  perfect  uniformity,  the 
number  of  oscillations  of  the  pendulum  of  the  clock  in  this  in- 
terval will  be  an  integer,  and  the  number  of  oscillations  of  the 
body  in  the  same  interval  will  be  less  by  one  complete  oscilla- 
tion than  that  of  the  pendulum  of  the  clock.  Hence  by  a 
simple  proportion  the  time  of  a  complete  oscillation  may  be 
found. 

The  coincidences  were  determined  in  the  following  manner : 
Certain  marks  made  on  the  two  pendulums  were  observed  by  a 
telescope  at  the  lowest  point  of  their  arcs  of  vibration.  The 
field  of  view  was  limited  by  a  diaphragm  t<j  a  narrow  aperture 
across  which  the  marks  were  seen  to  pass.  At  each  succeeding 
vibration  the  clock  pendulum  follows  the  other  more  closely, 
and  at  last  the  clock-mark  completely  covers  the  other  during 
their  passage  across  the  field  of  view  of  the  telescope.  After 
a  few  vibrations  it  appears  again  preceding  the  other.  The 
time  of  disappearance  was  generally  considered  as  the  time  of 
coincidence  of  the  vibrations,  though  in  strictness  the  mean  of 
the  times  of  disappearance  and  reappearance  ought  to  have 
been  taken,  but  the  error  thus  produced  is  very  small.  {Encijc. 
Met,  Figure  of  the  Eaith.)  In  the  experiments  made  in 
Ilartan  coal-pit  in  1854,  the  Astronomer  Royal  used  Kater's 
method  of  observing  the  pendulum.     {Phil,  Trans,,185Q,) 

The  value  of  T  thus  found  will  require  several  corrections. 
These  are  called  ''  Reductions."  If  the  centre  of  oscillation  does 
not  describe  a  cycloid,  allowance  must  be  made  for  the  altera- 
tion of  time  depending  on  the  arc  described.  This  is  called 
"the  reduction  to  infinitely  small  arcs."  If  the  point  of  sup- 
port be  not  absolutely  fixed,  another  correction  is  required 


220  A  STANDARD  OF  LENGTH.  [188.  J 

{Phil.  Trans.^  1831).  The  effect  of  the  buojancy  and  the 
resistance  of  the  air  must  also  be  allowed  for.  This  is  the 
"reduction  to  a  vacuum."  The  length  Z  must  also  be  cor- 
rected for  changes  of  temperature. 

The  time  of  an  oscillation  thus  corrected  enables  us  to  find 
the  value  of  gravity  at  the  place  of  observation.  A  correction 
is  now  required  to  reduce  this  result  to  what  it  would  have  been 
at  the  level  of  the  sea.  The  attraction  of  the  intervening  land 
must  be  allowed  for  by  Dr.  Young's  rule  {Phil.  Trans.,1819). 
"VVe  thus  obtain  the  force  of  gravity  at  the  level  of  the  sea, 
supposing  all  the  land  above  this  level  were  cut  off  and  the  sea 
constrained  to  keep  its  present  level.  As  the  sea  would  tend 
in  such  a  case  to  change  its  level,  further  corrections  are  still 
necessary  if  we  wish  to  reduce  the  result  to  the  surface  of  that 
spheroid  which  most  nearly  represents  the  earth.  (See  Carrib, 
Phil.  Trans.,  vol.  x.) 

There  is  another  use  to  which  the  experimental  determina- 
tion of  the  length  of  a  simple  '^'^ui\  "■  nt  pendulum  maybe 
applied.  It  has  been  ndopted  as  a  standard  of  length  on 
account  of  being  invariable  and  capable  at  any  time  of  recov- 
ery. An  Act  of  Parliament,  5  Geo.  lY.,  defines  the  yard  to 
contain  thirty-six  such  parts,  of  which  parts  there  are  39.1393 
in  the  length  of  the  pendulum  vibrating  seconds  of  mean  time 
in  the  latitude  of  London,  in  vacuo,  at  the  level  of  the  sea,  at 
temperature  62°  F.  The  Commissioners,  however,  appointed 
to  consider  the  mode  of  restoring  the  standards  of  weight  and 
measure  which  were  lost  by  fire  in  1834,  report  that  several 
elements  of  reduction  of  pendulum  experiments  are  yet  doubt- 
ful or  erroneous,  so  that  the  results  of  a  convertible  pendulum 
are  not  so  trustworthy  as  to  serve  for  supplying  a  standard  for 
length ;  and  they  recommend  a  material  standard,  the  distance, 
namely,  between  two  marks  on  a  certain  bar  of  metal  under 
given  circumstances,  in  preference  to  any  standard  derived 
from  measuring  phenomena  in  nature.   {Report,  1841.) 

All  nations,  practically,  use  this  simple  mode  of  determining 
the  length  of  the  standard  of  measure,  that  of  placing  two 
marks  on  a  bar,  and  by  a  legal  enactment  declaring  it  to  be  a 
certain  length. 


[139,  140.]  ELLIPTICITY  OF  THE  EARTH.  221 

139,  Form  of  the  Earth.     The  pendulum  t'urniehes  one  ot 
the  best  means  for  determining  the  form  of  the  earth. 
Let  a  =  the  equatorial  radius  of  the  earth ; 
h  =  the  semi-axis ; 
€  =  the  ellipticity  of  the  earth  ; 
then 


€  = 


a 


Let  m  =  ratio  of  the  centrifugal  force  at  the  equator  to  the 
force  of  gravity  at  the  same  place ; 
Iq  =  length  of  a  second's  pendulum  at  the  equator ; 
4o  =  the  length  of  a  second's  pendulum  at  the  poles ; 
then,  from  the  MScanique  Celeste,  tome  II.,  No.  34,  we  have 

The  value  of  m  is  ^-J^.  The  formula  for  the  length  of  the 
second's  pendulum  when  the  length  at  Paris  is  taken  as  unity,  is 

I  =  0.996823  +  0.00549745  sin^^, 
when  His  the  latitude  of  tjie  place.,.  See  Fuissant's  TraiU  d& 
Geod'Ssie,  P^g^  461. 

By  this  means  it  has  been  found  that  e  is  about  ^\j.  Bow- 
dich,  in  his  translation  of  the  Mecanique  Celeste^  p.  485,  re- 
marks, "  It  appears  that  the  oblateness  (e)  does  not  differ  much 
from  yJtj-,  and  may  possibly  be  a  little  more,  though  some  results 
give  a  little  less." 

140.  Torsion  Pendulum.  If  an  elastic  bar,  CD^  be  fixed 
at  one  end,  and  at  the  other  end  have 
two  weights,  A^  and  ^2?  rigidly  fixed  to 
it  by  means  of  the  cross  arm,  A^^  A^^  then 
if  the  arm  be  turned  into  the  position 
B^B^^  the  elastic  resistance  of  the  bar 
DC  will  cause  the  weights  to  move  back 


m:>b. 


to  -4iJ.2,  and  by  virtue  of  the  energy  of      '       ..-•"'' *^ ''*•'..       "^^ 
the  weights  at  that  point,  they  will  pass  p0  XOj 

that  position,  and  move  on  until  their     '  rio.  127. 

motion  is  arrested  by  the  action  of  the  elastic  resistance  of  the 


222  VIBEATIONS  OF  A  [141.] 

bar  ;  after  which  they  will  return  to  their  former  position,  thus 
having  a  motion  similar  to  that  of  the  common  pendulum.  This 
arrangement  is  called  a  torsion  jpendulum.  The  motion  will 
be  the  same  for  one  weight  as  for  two,  but  when  the  bar  DO  is 
vertical^  the  ai'ms  CA^  and  CA^  should  equal  each  other,  and 
the  weight  A-^  equal  A^. 

141.  To  find  the  force  necessary  to  twist  the  rod  DC  through 
a  given  angle^ 

Let  F=  the  force  at  A^  perpendicular  to  the  arm  CA^ ; 
a  :=  0A^=  OA,',  l  =  DC; 

a  =  AiGBi ;  /=  the  polar  moment  of  inertia  of  a  trans- 
verse section  of  the  bar  DG\  and 
G  =  the  coefficient  of  the  elastic  resistance  to  torsion. 

The  moment  of  the  twisting  force,  F,  will  be 

Ft\ 

and  the  moment  of  the  elastic  resistance  will  be  (see  Resistance 
of  Materials,  2d  Ed.,  p.  206), 


hence,  we  have 


Gl^--, 


Fa  =  (?/" ; 


A  F^  Gl% 

at 


The  weights  A^  and  ^.g  are  not  involved  in  this  problem. 

If  the  angle  be  measured  from  some  fixed  line  making  an 
angle  ^  with  the  neutral  position  of  J.i^2^  then  instead  of  a  we 
would  have  a^  —  ^,  and  the  last  equation  becomes 

If  the  force  be  reversed,  it  will  twist  the  bar  in  the  opposite 


[142. J  TORSION  PENDULUIVI.  223 

direction,  making  an  angle  a^  with  the  fixed  line  of  reference, 
and  we  would  have 

-J -y  (*-«.). 

Adding  these  equations  gives 

2  -J  =  -^  (oi  -  og). 

142,  Tofiiid.  the  time  of  an  oscillation. 
The  bar  CD  having  been  twisted  by  moving  the  bar  from  its 
normal  position  A1A2  into  the  position  B^B^^^  and  then  left  to 
itself,  it  is  required  to  find  the  time  of  moving  to  the  other 
extreme  position  B^B^.  We  will  neglect  the  mass  of  the  rod 
^1^2)  ^"d  that  of  the  bar  DC,  and  thus  simplify  the  solution, 
and  secure  an  approximate  result. 

Let  I^  =  the  moment  of  inertia  of  one  of  the  bodies,  Ai,  or 
A2,  in  reference  to  CD  as  an  axis ; 
6  =  2i  variable  angle  measured  from  the  neutral  posi- 
tion, AiA^ ;  and, 
considering  Fee  as  a  variable  moment,  producing  the  variable 
angle  6,  we  have  from  the  second  of  equations  (157),  and  the 
value  of  Fa  from  the  fii-st  equation  of  the  preceding  Article, 

Multiply  by  dd  and  integrate,  and  observing  that  for  0  =  a 
the  angular  velocity  is  zero,  we  find 

hence 


Integrating  again  gives 


224  DENSITY  OF  [14»A 

wliich,  between  the  limits  of  0  and  a,  gives 

which  is  the  time  of  half  an  oscillation ;  hence  the  time  for  a 
full  oscillation,  or  the  time  of  movement  from  B^  to  i?3,  will  be 

which  is  the  time  required.     The  times  are  isochronous  and  in- 
dependent of  the  amplitude. 

The  value  of  G  may  be  eliminated  by  substituting  its  value 
taken  from  the  last  equation  of  the  preceding  article.  Making 
the  substitution,  we  find 


—  TT  Y 


-  Oa) 


Fa 
from  which  I  and  /have  also  disappeared.     From  this  we  find 


Fa  =  I^{a^-a;){^ 


143.  To  find  the  density  of  the  earth. 

The  plan  of  determining  the  density  of  the  earth  by  means 
of  a  toi-sion  rod  was  first  suggested  by  the  Rev.  John  Mitchel. 
He  died  before  he  was  able  to  make  the  experiment,  but  the 
plan  was  executed  by  Mr.  Cavendish,  who  published  the  result 
in  the  PhiL  Trans,  for  1798.  Subsequent  to  1837,  Mr.  Bailey, 
at  the  request  of  the  Astronomical  Society  (England),  made  a 
new  determination  of  the  result.  He  made  upwards  of  2,000 
experiments  with  balls  of  different  weights  and  sizes,  and  sus- 
pended in  a  variety  of  ways,  a  full  account  of  which  is  given 
in  the  Memoirs  of  the  Astronomical  Society^  Yol.  xiv.  We 
give  here  only  some  of  the  more  prominent  features  of  the  f5x- ' 
periment. 

The  torsion  rod  DO  was  very  small,  so  that  it  could  be  easily 
twisted.     Two  small  balls,  J.i,  ^2?  were  suspended  from  the 


[143.] 


THE  EARTH. 


225 


torsion  rod  by  a  light  cross-bar.  Two  large  balls,  Ei  E2,  were 
placed  oil  a  plank  which  turned  about  a  point  O  directly  under 
C,  and  the  whole  so  arranged  that  the  centres  of  gravity  of  the 
four  balls  were  in  the  same  horizontal  plane.     The  apparatus 


Fio.  128. 

was  inclosed  in  a  small  room  so  as  to  exclude  currents  of  air, 
and  the  weights  E^  and  E2  were  moved  into  the  desired  posi- 
tions from  the  outside  of  the  room  by  means  of  mechanism  ex- 
tending into  the  room. 

The  weights  E^  and  E^  were  first  placed  nearly  at  right 
angles  with  the  rod  AiA^  when  the  latter  would  assume  some 
neutral  position  as  Ca.  The  balls  E1E2  were  then  brought 
quite  near  to  the  small  ones,  j^i,  JB^^  when  the  attraction  of  the 
former  drew  the  latter  from  their  neutral  position,  and  they 
oscillated  about  some  position  of  equilibrium  as  B^  B^.  The 
angle  aCB^  was  observed,  and  also  the  time  of  the  oscillation 
about  the  position  CBz* 

The  balls  were  then  changed  to  the  position  i^/^,  making 
the  angle  aCF^  as  nearly  equal  as  possible  to  aCE2 ;  but  it  was 
found  that  the  line  Ca  did  not  always  bisect  the  angle  E2  CFz^ 
but  the  mean  of  many  readings  was  taken  as  the  most  probable 
value.     The  angle  E^CF^  will  be  oti  —  03,  given  in  the  preced- 

ino'  Article. 

J?  « 

It  is  proved,  by  the  law  of  attraction,  that  the  attraction  of  a 
homogeneous  sphere  is  the  same  as  if  its  entire  mass  were  con- 
centrated at  the  centre  of  the  mass,  and  varies  inversely  as  the 
square  of  the  distance  from  the  centre. 

Let  M—  the  mass  of  one  of  the  large  balls ; 


m  = 


a     u     u     u 


16 


small  balls ; 


220  DENSITY  OF  [143.1 

D  =  the  distance  between  their  centres ;  and 
fjb    =  the  attraction  of  a  sphere  whose  mass  is  unity  upon 
another  unit  when  the   distance   between   theii 
centres  is  unity ; 
then  the  force  of  attraction  of  the  mass  M  upon  m  will  be 

Mm 


H' 


ly 


and  this  is  the  value  of  i^in  the  last  equation  of  the  preceding 
Article.     But  there  being  t^ 
this  attractive  force  will  be 


Article.     But  there  being  two  balls  in  this  case,  the  moment  of 


^    Mm 

which  (by  neglecting  the  attraction  of  the  large  ball  and  plank 
npon  the  rod  CB^^  and  of  the  plank  upon  the  small  balls), 
equals  the  second  member  of  the  last  equation  of  the  preceding 
Article.     Hence, 

Mm 


2/^-]^^  —  A(ai  -  ^)(^) 


Let  ^be  the  mass  of  the  earth.  It  its  radius,  and  ^  the  force 
of  gravity,  then 

E 


Eliminating  /x,  and  making  I^  =  m  (c?  +  f  /^),  we  have 


The  density  of  the  earth  is  thus  reduced  to  the  determination 
of  the  Oi  —  os  between  its  two  positions  of  equilibrium  when 

*  In  Bailey's  experiments,  the  value  used  was 
E 


5'  =  At;^[l-2e  +  (fm-6)  cos^aJ  ; 


in  whicli  c  is  the  ellipticity  of- the  earth,  m  the  ratio  of  the  centrifugal  force  at 
the  equator  to  equatorial  gravity,  and  \  the  latitude  of  the  place. 


[144.]  THE  EARTH.  227 

under  the  action  of  the  masses  iu  their  alternate  positions,  and 
the  time  T  of  oscillation  of  the  torsion  rod.  To  observe  these, 
a  small  mirror  was  attached  to  the  rod  at  O^  with  its  plane 
nearly  perpendicular  to  the  rod.  A  scale  was  engraved  on  a 
vertical  plate  at  a  distance  of  108  inches  from  the  mirror,  and 
the  image  of  the  scale  formed  by  reflection  on  the  mirror  was 
viewed  in  a  telescope  placed  just  over  the  scale.  In  this  way 
an  angle  of  one  or  two  seconds  could  be  read. 

The  final  result  was  that  the  mean  density  of  the  earth  is 
5.6747  times  that  of  distilled  water  at  its  maximum  density. 

144.  Problem.     If  the  earth  were  a  Jiomogeneous  aph&re^  at 
what  point  in  the  radius  must  it  he  struck^  and  what  momentum 
must  it  receive^  that  it  shall  have  a  velocity  of  translation  of 
V  and  of  rotation  of  to? 
Let  M  =  the  mass  of  the  sphere, 
jff  =  its  radius, 
ki  =   V^B  =  the   principal    radius   of  gyration.     (See 

Example  4,  page  174),  and 
a   =  the  distance  from  the  centre  to  the  point  where  the 
impulse  is  applied. 
The  momentum  must  be 

JfF, 

;;\'herever  the  blow  is  applied.  The  moment  of  an  impulse 
being  the  same  as  the  moment  of  the  momentum,  we  have,  ac- 
cording to  equation  (155), 

moment  of  the  momentum, 

0)  = -. ; 

moment  of  tnertia 

_  MV.a 

_  Va_ 


228  ANGULAR  VELOCITY.  [145.  | 

TJie  angular  velocity  of  the  earth  per  hour,  ia 

27r  ^ 

and  tlie  linear  velocity  in  the  orbit  is 

68,000  miles  per  hour  nearly ; 

••  ""- 2,040,000  ■^'• 
Letting  R  =  4,000  miles,  we  have 

«  =  24  miles  nearly. 

145,  Problem.  A  homogeneous  disc  has  a  motion  of  trans- 
lation and  of  rotation  entirely  in  its  own  jplane^  when  sicddenh, 
any  point  in  the  disc  becomes  fixed  ;  required  the  angidar  velo- 
city about  the  fixed  point. 

Let  V  =  the  velocity  of  translation  of  the  centre  of  the  disc : 

CO  =  the  'angular  velocity  about  the  centre ; 

jp  =  the  perpendicular  distance  between  the  fixed  point 

and  the  line  of  motion  of  the  centre  at  the  instant 

that  the  point  becomes  fixed  ; 

r  =  the  distance  between  the  fixed  point  and  the  ceLtre 

of  the  disc  ; 
ki  =  the  principal  radius  of  gyration  ;  and 
«i=  the  angular  velocity  about  the  fixed  point. 
Then 


moment  of  the  momentum 
moment  ofiner 

Mh^(o-\-MVp 


^  moment  of  inertia 


_l\^w-\-Vp 
~    h^  +  r"    ' 

If  F=  0,  we  have  '     • 


a>i  = 


1146.  J  OF  FINITE  MASSES.  229 

If  the  centre  becomes  fixed,  we  have  ^  =r  0,  and  r  =  0,  and 

(Oi  =  ft). 

1 46.  I*EOBLEM.  A  sphere  whose  radius  is  R  has  an  angular 
velocity  (o^  and  gradually  contracts  until  its  radius  is  r  /  re- 
quired the  final  angxdar  'velocity. 

We  will  assume  that  the  body  remains  homogeneous  through- 
out, and  that  there  is  no  change  of  temperature,  and  that  the 
change  of  volume  is  due  simply  to  the  mutual  attraction  of  the 
particles  for  each  other,  which  is  supposed  to  draw  them  towards 
the  centre.  Then  will  the  moment  of  the  momentum  be  constant. 

Let  0)1  be  the  required  angular  velocity  ;  then  we  have  from 
equation  (155) 

Qa  =  (oM.\  E"  =  ft)iJ/i.f  /^  ; 

•••  «i  =  ^  ^- 

Problem. — A  spherical  homogeneous  mass  m,  radius  r,  contracts  by  the 
mutual  attraction  of  its  particles  to  a  radius  nr  ;  if  the  work  thus  expended 
be  suddenly  changed  into  heat,  how  many  degrees  F.  will  the  temperature 
of  the  mass  be  increased,  its  specific  heat  being  5  and  the  heat  uniformly 
disseminated  ? 

Consider  the  earth  as  a  homogeneous  sphere,  mass  Jf,  radius  B,  S  its  den- 
sity compared  with  water,  g  the  acceleration  on  the  surface  of  the  earth 
due  to  gravity,  g'  that  on  the  surface  of  m.  Then,  since  attractions  are 
directly  as  the  masses  and  inversely  as  the  square  of  the  radii,  we  have  g'  = 

jj.*  -J  g.     Initially,  the  force  within  the  sphere  varies  directly  as  the  distance 

p  from  the  centre,  but  during  contractions  as  the  inverse  squares;  hence  the 

acceleration  at  a  distance  a?  of  a  particle  originally  at  p  will  be  -tm  •  —T^g .-  . 

_1 

^.     The  mass  of  a  spherical  shell,  radius  p,  is  dm  =  {m  -i-  \  irr^)  iirp^dp  = 

-j-  p^dp.     The  force  equals  the  mass  into  the  acceleration,  and  this  by  dx 
will  be  an  element  of  work  ;  hence  the  total  work  will  be 

3m«   if      rr  rnp  p^  Zm^  B^l-n 

Dividing  by  J"  (Joule's  mechanical  equivalent  of  heat),  also  by  the  weight 
of  a  sphere  of  water  eqnal  in  volume  to  that  of  the  given  sphere,  or  -^  .  — ^, 

and  also  by  the  specific  heat  of  the  body,  we  finally  have  T  =  t— r-ir  • 

— Problem  by  the  Author  in  Mathematical  Visitor^  July,  1880. 


CHAPTER    XI. 

GENERAL   EQUATIONS    OF   MOTION. 

147.  D'Alembert's  Principle. — A  body  is  a  collection  of 
material  particles  held  together  bj  a  force  exerted  by  each  parti- 
cle upon  the  others.  Having  deduced  the  laws  of  action  for 
forces  acting  upon  a  single  particle,  the  direct  process  for  d(5- 
termining  the  effect  of  forces  upon  a  body  of  finite  size,  appears 
to  be  to  consider  all  the  forces  which  act  upon  each  particle 
separately,  including  the  mutual  actions  and  reactions  of  the  par- 
ticles, thus  establishing  equations  for  each  particle  of  the  body, 
and  then  to  eliminate  the  terms  involving  the  actions  and  reac- 
tions. But  the  latter  are  generally  unknown.  Yarious  expedients 
were  resorted  to  by  the  ancient  mathematicians  to  reach  the  re- 
sulting equations,  but  the  principle  announced  by  D'Alembert 
greatly  simplified  the  operations,  and  in  many  cases  reduced  the 
establishment  of  the  equations  to  Statical  principles.  See 
Whewell's  History  of  the  Inductive  Sciences,  Yol.  1.,  p.  365. 

The  forces  which  produce  the  motion  of  a  body  may  be 
applied  at  only  a  few  points,  and  yet  produce  motion  in  every 
particle  of  it.  These  forces  are  called  impressed  forces.  If  we 
consider  the  particles  as  separated  from  each  other,  and  forces 
applied  to  them  which  will  produce  the  same  motion  that  they 
had  when  in  the  body,  the  latter  forces  are  called  effective 
forces.  The  effective  forces  will  then  produce  the  same  effect 
upon  the  body  as  the  impressed  forces.  D'Alembert's  principle 
consists  in  this,  that  if  a  system  of  forces^  equal  and  opposite 
to  the  EFFECTIVE  FORCES,  act  upou  a  hody^  they  will  he  in  equili- 
hrium  with  the  impressed  forces. 

In  this  principle  no  assumption  has  been  made  in  regard  to 
the  character  of  the  mutual  actions  and  reactions  between  the 
particles,  and  hence  it  is  applicable  to  flexible  bodies  and  fluids, 
as  well  as  to  solids.     It  is  equivalent  to  assuming  that  the  forces 


i 


[148.]  GENERAL  EQUATIONS   OF  MOTION.  231 

within  a  body  constitute  a  system  which  are  in  equilibrium 
among  themselves. 

148,    To  FIND  THE   GENERAL    EQUATIONS  OF    MOTION   OF   A  BODY. 

Let  a?i,  yi,  2i,  be  the  coordinates  of  a  particle  whose  mass  is  m^ ; 
Xi^  JTi,  Zi,  the  impressed  forces  parallel  to  the  respective  axes 
acting  upon  the  particle ;  and  a  similar  notation  for  the  other 
particles.  The  measure  of  the  accelerating  force  parallel  to  the 
axis  of  X  will  be 

and  if  a  force  equal  and  opposite  to  this  act  upon  the  particle 
there  will  be  equilibrium ;  hence  we  have 

Similarly, 

and  similar  expressions  for  all  the  other  particles  of  the  body. 
But  if  JX,  X  Y,  XZ,  be  the  sum  of  the  respective  axial  com- 
ponents of  the  impressed  Jbrces,  then 

XX  =  Xi  +  Xa  +  X3  +  etc. ; 

and  similarly  for  the  (Others.  Hence,  if  771  be  the  mass  of  any 
particle  whose  coordinates  are  x,  y,  2,  at  the  time  ^,  we  have 
according  to  D'Alembert's  principle 

and  similarly  for  the  others. 

Taking  the  moments  of  the  forces  in  reference  to  the  axis  of 
«,  we  have,  in  precisely  the  same  way,  the  equation 

and  similarly  for  the  other  axis.     Hence,  we  have  the  follow 
ing  six  equations : — 


2B2 


aBNERAL  EQUATIONS  OF 


%Z-X.n%=^- 


ri«.j 


(164) 


fPy\ 


%{Zy  -    Yz)-X  (my  II  -  mz  g)  =  0  ; 

-ZiXz  -  Zx)-^  (mz  g  -  ,nx  J)  =  0 ;     j-    (165) 

X{Yx  -Xy)-X  (m*g  -  my  5)  =  0. 

Let  a?!,  ?/i,  ^,  be  the  coordinates  of  any  particle  of  the  bo(iy 
referred  to  a  movable  system  whose  origin 
remains  at  the  centre  of  the  mass,  and  whose 
axes  are  parallel  to  the  fixed  axes,  and 
X,  y,  i,  the  coordinates  of  the  centre  of  the  mass  referred 
to  the  fixed  origin ; 
then  we  have 

X  =  x  +  Xi\ 

Smx  =  Xmx  +  Xmxi ; 

Bnt  tlie  origin  of  the  movable  system  being  at  the  centre  of 
the  mass,  we  have,  frpm  equations  {71a)  or  (84^), 

^msoi  =  0 ; 

and  the  last  of  the  preceding  equations  becomes 
_     d^x       _,     d^x 


[14a] 


MOTION. 


233 


since  -r?^-   is  a  common  factor  to  all  the  particles  m\   but 
^m  =  J/"; 

and  similarly  for  the  others.      Hence,  equations  (164)  become 


(16S) 


Similarly,  in  equations  (165),  we  have 


d^je 


(^^ 


But  y  and  i  are  common  factors  in  their  respective  terms, 
therefore  the  expression  becomes 


but, 


^^yi  =  0 ; 


hence,  we  finally  have 


d?2        ,._  <^      ^        ^2i 

and  similarly  for  other  terms.     In  this  way  the  first  of  eqna 
tious  (165)  becomes 


234 


GENERAL  EQUATIONS  OF 


[148.J 


Multiply  the  third  of  equations  (166)  by  y,  the  second  by  s, 
subtract  the  latter  from  the  former,  and  we  have 


.^d}z 


^y-i:%-^^ 


.^y 


df 


^Zy-'SYz', 


which,  substituted  in  the  preceding  equation,  gives 

Dropping  X  before  JT,  T,  Z,  and  letting  those  letters  repre- 
sent the  total  axial  components  upon  the  entire  body,  and 
Zy^  —  J^i,  etc.,  the  resultant  moments  of  the  applied  forces, 
we  have  the  six  following  equations : 


*s= 

X; 

'S= 

T; 

*s= 

Z; 

d^Xy 


(167) 


(168) 


Equations  (167)  do  not  contain  the  coordinates  of  the  point 
of  application  of  the  forces,  hence,  the  motion  of  translation 
of  the  centre  of  a  mass  is  independent  of  the  point  of  ajpplicor 
tion  of  the  force  or  forces  /  or,  in  other  words,  it  is  ind-epen- 
dent  of  the  rotation  of  the  mass. 

Equations  (168)  do  not  contain  the  coordinates  of  the  centre 
of  the  mass,  and  being  the  equations  for  rotation,  show  that 
the  rotation  of  a  mass  is  independent  of  the  translation  of  its 
centre. 


1149,150.]  MOTION.  230 

These  equations  are  sufficient  for  determining  all  the  circum- 
stances of  motion  of  a  free  solid.  In  theii'  further  use  the 
dashes  and  subscripts  will  be  omitted. 

149.  If  X,  Y,  Z,  are  zero,  we  have 

and  similarly  for  the  others.      Transposing,  squaring,  adding 
and  extracting  the  square  root,  give 


^'^'V ie =  V6^,«+6i^+6;;    (169) 

which,  being  constant,  shows  that  the  motion  of  the  centre  of  the 
mass  is  rectilinear  and  uniform. 
This  is  the  general  principle  of  the  conservation  of  tub 

OBNTBE  OF  GRAVITY. 


CONSERVATION   OF   AREAS. 

150.  The  expression, 

xdy  —  ydx^ 

is,  according  to  Article  112,  twice  the  sectoral  area  passed 
over  by  the  radius  vector  of  the  body  in  an  instant  of  time. 
Hence,  if 

^{rnxdy  —  m,ydx)  =  dAi ; 

differentiating,  we  find 

^/       ^y  d^x\        d^Ai 

If  there  are  no  accelerating  forces  m  -^  =  0 ;  and  similarly 
for  the  others ;  hence 


df  -^' 

.-.  Ai  =  Git;        Ai=:c^; 
1.5 

A,  =  c^; 

(170) 

i'.yL^  r-  1    V ' 

.X^' 


236  CONSERVATION  OP  [ISl.j 

the  initial  values  being  zero.  These  are  the  projections  on  the 
coordinate  planes  of  the  areas  swept  over  by  the  radius  vector 
of  the  body.  They  establish  the  principle  of  the  conservation 
OF  AREAS.     That  is, 

In  any  system  of  bodies,  moving  without  accelerating  forces 
and  having  only  mutual  actions  upon  each  other,  the  projections 
on  any  plane  of  the  areas  swept  over  hy  the  radius  vector  are 
proportional  to  the  times. 


CONSERVATION   OF    ENERGY. 

151.  Multiply  equations  (167)  by  dx,  dy,  dz,  respectively, 
add  and  integrate,  and  we  have 

Mv''  -  Mv^  =  ^f{Xdx  +  Tdy  +  Zdz) ; 

and  for  a  system  of  bodies,  we  have 

^X{Mv^)  -  i^{Mv,')  =  SfiXdx  +  Tdy  +  Zdz).     (171) 

The  second  member  is  integrable  when  the  forces  are  directed 
towards  fixed  centres  and  is  a  function  of  the  distances  between 
them. 

Let  a,  b,  c  be  the  coordinates  of  one  centre, 
<^i)  ^15  ^15  of  another,  etc. ; 
X,  y,  3,  the  coordinates  of  the  particle  m^ ; 
r,  7*1,  etc.,  be  the  distances  of  the  particle  from  the  res- 
pective centres; 
I]  I[,  etc.,  be  the  forces  directed  towards  the  respective 
centres ; 
then,  resolving  the  forces  parallel  to  the  axes,  we  have 

X=  Faos  a  }-  JFl  cos  a  +  etc. 

^a  —  X       ^a^  —  X 

=  F h  F^ h  etc. ; 

r  r 

r                   r 
Z  =  F +  Fi- 4-  etc. 


[151.1  ENERGY.  237 

Multiplying  by  dx,  dy,  dz,  respectively,  and  adding,  we  have 
Xdx  +  Ydy  +  Zds  =  f\  ^^^^dx  +  -^^dy  +  ^-^dz  | 

^jr\^J^-d.'+^-^^dy  +  '^^'d.\ 

I     7*1  n  ^1       ' 

+  etc. 
Bnt  7^  =  {a-  xf  +  {h  -yf  +  {c  -  zf ;  (172) 

and  by  differentiating,  we  find 


similarly, 


a-x           h-y           c-  Zj 
dr  =  — dx  — ^y az  ; 


,  a^  —  x,        \  —  y  ^1— ^T 

dvx  —  — dx -dy dz  ; 

r^  n  n 


etc.,  etc.,  etc. 

These  substituted  above,  give 

Xdx  +  Ydy  ^  Zdz—  —  Fdr  -  F^dr^  -  F/ir^  -  etc. 

Therefore,  if  F^  etc.,  is  a  function  of  r,  etc.,  and  /Lt,  /^i,  etc., 
the  intensities  of  the  respectiye  forces  at  a  distance  unity  from 
the  respective  centres,  or 

F=fJLct>(r); 

F,  =  fJL,<l>{r,)',         ^ 

etc.,      etc. ; 

the  second  member,  and  hence  the  fii-st,  will  be  integrable. 

In  nature,  if  a  particle  m  attracts  a  particle  m^,  the  particle 
7ni  will  attract  m,  each  being  a  centre  of  force  in  reference  to 
the  other,  and  both  centres  will  be  movable  in  reference  to  a 
fixed  origin.  But  one  centre  may  be  considered  fixed  in  ref- 
erence to  the  other,  and  consequently  the  proposition  remains 
true  for  this  case. 

The  second  member  of  equation  (171)  being  integrated  be- 
tween the  limits  a?,  y,  z  and  x^,  y^,  z^^  we  have 

\:E{M^-  mMvo')  =  2m9  (^,  2/1  ^)  -^M9  (^0,  yo,^o).  (173) 

Hence,  the  gain  or  loss  of  energy  of  a  system^  subject  to 

forces  directed  towards  fixed  centres  and  winch  are  functions 

of  the  distances  from  those  centres,  is  independent  of  the  path 


2SS  CONSERVATION  OF  [151.  j 

described  hj  the  bodies,  and  dejpends  only  upon  the  position  left 
ami  arrived  at  by  the  bodies,  and  the  intensities  of  the  forces 
at  a  unifs  distance  from  the  respective  centres. 

Therefore,  when  the  system  returns  to  the  initial  position,  or 
to  a  condition  equivalent  to  the  original  one,  the  vis  viva  will 
be  the  same. 
From  equation  (171)  we  have 

The  gain  or  loss  of  energy  in  passing  from  one  position  to 
another  eqitals  the  work  done  by  the  impressed  forces. 

Let  TFo  =  the  work  done  by  the  impressed  forces  in  passing 
from    some  definite  point  (a?o,  Vo,  ^o?)   to  an- 
other definite  point  (a?i,  ?/i,  ^i,);  and 
W  =  the  work  done  by  the  same  forces  in  passing  from 
the  first  point  to  any  point  (a?,  y,  ^,) ; 
then  from  equation  (171),  we  have 

i:E{3£v,')-i:E{Mv')  =W', 
and  subtracting  the  latter  from  the  former,  gives 

i^{IM)  -  i:E{Mv^^)  =  TTo  -  TT; 
or,  by  transposing, 

^:E{Mv')  +  TF  =  i:2{Mv,^)  +  W, ; 

in  which  the  second  member  is  constant. 

The  first  term  of  the  first  member  is  the  kinetic  energy 
w^hich  the  system  has  at  any  point  of  the  path,  and  the  second 
term  is  the  work  which  has  been  done  by  the  forces  upon  the 
body  and  has  become  latent,  or  potential ;  hence,  in  such  a 
system  the  sum  of  the  kinetic  and  potential  energies  is  con- 
stant. 

This  is  the  principle  of  the  conservatiotq"  of  energy  in 
theoretical  mechanics.  This  term  has  been  extended  so  as  to 
include  the  principle  of  the  transmutation  of  energy  as  estab- 
lished by  physical  science. 

If  a  portion  of  the  universe,  as  the  Solar  System  for  in- 
stance, be  separated  from  all  external  forces,  tlie  sum  of  the 
kinetic  and  potential  energies  will  remain  constant,  so  that  if 
the  kinetic  energy  diminishes,  the  potential  increases,  and  the 


[151.1  EXERQY.  239 

converse.  If  external  forces  act,  the  potential  and  kinetic 
energies  may  both  be  increased. 

To  be  more  specific,  suppose  that  the  earth  and  sun  consti- 
tute the  system,  the  sun  being  considered  the  centre  of  the 
force.  The  velocity  of  the  earth  will  be  greatest  when  nearest 
the  sun,  and  will  diminish  as  it  recedes  from  it.  While  reced- 
ing, the  amount  of  work  done  against  the  attractive  force  of 
the  snn  will  be 

in  which  -3f  is  the  mass  of  the  earth,  Vi  the  maximum  velocity, 
and  V  the  velocity  at  any  point.  The  second  member  is  nega- 
tive because  the  first  member  is. 

When  the  earth  is  approaching  the  sun  the  velocity  is  in- 
creased and  the  living  force  is  restored,  and  the  kinetic  added 
to  the  potential  energy  is  constant. 

Again,  if  a  body  whose  weight  is  W  be  raised  a  height  h, 
the  work  which  has  been  done  to  raise  it  to  that  point  is  Why 
and  in  that  position  its  potential  energy  is  Wh.  If  it  falls 
freely  through  that  height  it  will  acquire  a  velocity  v=zV  2gh, 

^g 
which  is  the  kinetic  energy. 

If  the  same  body  fall  through  a  portion  of  the  height,  say 
Ai,  its  kinetic  energy  will  be  Wh^  =  iMv^,  and  the  work  which 
is  still  due  to  its  position  is  W  {h—h<^,  which,  at  that  instant, 
is  inert  or  potential. 

It  is  found,  however,  that  in  the  use  of  machines  or  other 
devices,  by  which  work  is  transmitted  from  one  body  to  another, 
all  the  work  stored  in  a  moving  body  cannot  be  utilized.  Thus, 
in  the  impact  of  non-elastic  bodies  there  is  always  a  loss  of 
living  force.     (See  Article  32.) 

This,  so  far  as  theoretical  mechanics  is  concerned,  is  a  loss, 
and  is  treated  as  such,  and  until  modern  Physical  Science 
established  tJie  correlation  of  forces  it  was  supposed  to  be 
entirely  lost.  But  we  know  that  in  the  case  of  impact  lieat  is 
developed,  and  Joule  determined  a  definite  relation  between 
the  quantity  of  heat  and  the  work  necessary  to  produce  it,  and 
called  the  result  the  mechanical  equivalent  of  heat.     Further 


240 


ROTARY  MOTIONS. 


[152.] 


investigations  show  that  in  every  case  of  a  supposed  loss  of 
energy,  it  may  be  accounted  for  in  a  general  way  by  the 
appearance  of  energy  in  some  other  form.  It  is  impossible  to 
trace  the  transformation  of  energy  as  it  appears  in  mechanical 
action,  friction,  heat,  light,  electricity,  megnetism,  etc.,  and 
prove  by  direct  measurements  that  the  sum  total  at  every 
instant  and  with  every  transformation  remains  rigidly  con- 
stant; but  by  means  of  careful  observations  and  measurements 
nearly  all  the  energy  in  a  variety  of  cases  has  been  traced  from 
one  mode  of  action  to  another,  and  the  small  fraction  which 
was  apparently  lost  could  be  accounted  for  by  the  imperfec- 
tions of  the  apparatus,  or  in  some  other  satisfactory  manner ; 
until  at  last  the  principle  of  the  conservation  of  energy  is 
recognized  to  be  a  law  as  universal  as  that  of  the  law  of  gravi- 
tation. The  exact  nature  of  molecular  energy  which  manifests 
itself  in  heat,  chemical  affinity,  etc.,  are  unknown,  but,  accord- 
ing to  the  general  law,  all  energy  whether  molecular  or  of 
finite  masses,  is  either  kinetic  ov  potential, 

COMPOSITION   OF   KOTARY   MOTIONS. 

152.  Take   the   origin  of   coordinates  at  the  fixed  point 
when  there  is  one,  or  at  any  point  of  the  axis  of  rotation 
when  the  body  is  free.     Conceive  three  cones  having  a  com- 
mon vertex  at  the   origin  of  co- 
ordinates and   each    tangent   re- 
spectively to  the  coordinate  planes; 
then  will  the  angular  velocity  of 
that  element  of  the  cone  which  for 
the  instant  is  in  contact  with  the 
plane  xy,  considered   as  rotating 
about  the  axis  of  z^  be  defined  as 
the  angular  velocity  of  the  hody 
about  that  axis  ;  and  similarly  foi* 
the  other  axes.     Let  g?^,  gj^,  od^,  be  the  angular  velocities  of 
the  hody  about  the  respective  axes  x,  y,  z.     If  ca^  considered 
as  infinitesimal,  be   the  actual  velocity  of   a  particle  in  the, 
plane  xy^  positive  from  x  towards  y,  p  =  Oc^  ca  =  pcog ;  then 
cd  =  pGDg  cos  (180°  —  acd)  =  —  &?«  •  p  sin  XOc  =  —  cw^y, 
da  =  pcog  sin  (180°  —  acd)  =  go^  >  p  cos  XOc  =  a?^. 


a 


;:/in, 


Fig.  127. 


[152.]  ROTARY  MOTION.  241 

Similarl}^  in  regard  to  the  axis  of  y^  we  have 

and  in  regard  to  a?, 

When  all  these  rotations  take  place  at  the  same  time,  we 
have,  by  adding  the  corresponding  velocities,  the  several  velo- 
cities along  the  axes 


'di 


=  ft)y3  —  (Ozy 


dy 


(174) 


dz 
dt 


=  eoxy  —  (OyX. 


The  particles  on  the  instantaneous  axis  have  no  velocity  in 
reference  to  the  movable  origin,  hence 


dx 


dt       ' 


^2  _n. 


/.  (OyS  —  (Oty  =  0,     cjgX  —  Oa^  =  0,     co.j.y  —  w^  =  0 ;  (175) 

which  are  the  equations  of  a  straight  line  through  the  origin, 
and  are  the  equations  of  the  instantaneous  axis.  Let  a,  y8,  7,  be 
the  angles  which  it  makes  with  the  axes  a?,  y,  3,  respectively, 
then  (Anal,  Geom.), 


cos  a  =     .    o   .        i~ i ;    cos  yS  =   ,/  "T  .        o  .       2  > 

V  ft)/  +  ft)/  +  ft)/  ^        V  ft)a.2  +  ft)  2  +  0,2 ) 


cos  7 


4/, 


Wa:    +  O)/  + 


To  determine  the  angular  velocity  of  the  body,  take  any 
point  in  a  plane  perpendicular  to  the  instantaneous  axis.  Let 
the  point  be  on  the  axis  of  a?,  and  from  it  erect  a  perpendicular 
to  the  instantaneous  axis,  and  we  have 


j>=aj6ina=:a;Vl  —  cos'"^  a 


24-2  ROTARY  MOTIONS.  [153,  154.] 

For  this  point  y  =  0  and  s  =  0  in  equations  (174),  and  we 
find  for  the  actual  velocity, 

V=  -^ =  X  V(Dy'  +  0)/; 

and  hence 

V 


which  represents  the  diagonal  of  a  rectangular  parallelopipe- 
don,  of  which  the  sides  are  ©a.,  cd^,  o),. 

153,  Moments  of  rotation  of  the  centre  of  the  mass  about 
the  fixed  axes. 

Multiply  the  second  of  equations  (167)  by  2,  the  third  by  y, 
subtract  the  former  from  the  latter,  and  we  have 

Treating  the  equations  two  and  two  in  this  manner,  dropping 
the  dashes,  and  substituting  Zi,  M^^  iV^i,  for  the  second  mem- 
bers, w^e  have 


These  equations  are  of  the  same  j^<?rm  as  equations  (168). 

MOTION   OF    A   BODY   DURING   IMPACT. 

154,  Motion  of  the  centre  of  the  masses.  The  second  mem- 
bers of  equations  (167)  are  the  accelerating  forces.  If  any  two 
of  the  bodies  collide,  they  being  free  in  other  respects,  the 
action  of  one  body  upon  the  other  is  equal  and  opposite  to  that 
of  the  latter  upon  the  former;  hence,  in  regard  to  the  system 
they  neutralize  each  other,  and  the  motion  of  the  centre  of  the 
Diasses  will  be  unaffected  by  the  collision.     If  there  are  no 


[155,  150.]  CONSTRAINED   MOTION.  243 

accelerating  forces  the  velocity  of  the  centre  will  be  uniform 
and  in  a  straight  line,  as  shown  in  Article  14Q^  '■'-'' 

To  find  the  velocity  of  the  bodies  after  impact  requires  a 
knowledge  of  their  physical  constitution.  See  Articles  28  and  29. 

155.  T/ie  motion  of  rotation  of  the  centre  of  the  entire  mass 
about  the  origin  will  also  be  unaffected  by  the  collision,  when 
the  bodies  are  acted  upon  by  accelerating  forces ;  for,  the  mo- 
ments of  the  forces  due  to  the  collision  will  neutralize  each 
other,  and  the  second  members  of  equations  (177)  will  contain 
only  the  applied  forces. 

This  would  be  illustrated  by  the  impact  of  two  asteroids,  or 
in  the  bursting  of  a  primary  planet. 

But  the  rotation  produced  in  each  body  about  the  centre  of 
its  mass  depends  upon  the  moments  of  the  forces  applied  to  the 
body,  and  hence,  upon  the  moment  of  the  momentum  produced 
by  the  impact. 

CONSTEAINED    MOTION. 

156.  General  equations  of  rotation  about  a  fixed  jpoint. 
Take  the  origin  of  coordinates  at  the  fixed  point.     For  this 

case  equations  (164)  vanish.     Substitute  in  (165),  the  values  of 

-T7i ,  etc.,  obtained  from  (174),  and  reduce.     We  have 
d^x        doiy         d(o^  * 

and  similarly  for  the  others. 

Let  Z,  J/",  i\r,  be  substituted  for  the  last  term  respectively  of 
equations  (165),  and  substituting  the  above  values  in  the  last  of 
these  equations,  we  find 

-J  ^m{a?  +  y^)  +  (D^y  Sm{x^  —  'f) 
-  (o),'  -  f^y')  Xmxij  -  (^  +  a,,a,,)  Xmyz    )■  =  N.     (178) 


2M  PRINCIPAL  [157.J 

The  other  two  equations  may  be  treated  in  the  same  manner. 
But  they  are  too  complicated  to  be  of  use.  Since  the  position 
of  the  axes  is  arbitrary',  let  them  be  so  chosen  that 

Smxy  =  0,         ^?nx3  =  0,         Zmys  =  0 ;       (179) 
in  which  case  the  axes  are  called  princijpal  axes  /  and  we  will 
show  in  the  next  article,  that,  for  every  point  of  a  body,  there 
are  at  least  three  principal  axes,  each  of  which  is  perpendicular 
to  the  plane  of  the  other  two. 

Let  ^1,  yi,  ^1,  be  the  principal  axes,  having  the  same  origin 
as  the  fixed  axes,  and 

A  =  ^m{y^-\-2^),  the  moment  of  inertia  of  the  body  about  a^; 
B  —  Zin{zi^  +  x^)^  moment  about  yi ; 
O  —  2^m{xi^  +  y^),  moment  about  %  ; 

also  let  a>i,  co^,  ewa,  be  the  angular  velocities  about  the  respective 
axes  Xi,  2/1,  and  ^j,  and  substituting  these  several  quantities  in 
(178),  we  have 


(180) 


These  are  called  Euler's  EqTiations. 

The  origin  of  coordinates  may  be  taken  at  the  centre  of  the 
mass,  and  as  the  rotation  about  that  point  is  the  same  whether 
that  point  be  at  rest  or  in  motion,  as  shown  at  the  bottom  of 
page  234,  equations  (ISO)  are  applicable  to  the  rotation  of  a 
free  body  when  acted  upon  by  forces  in  any  manner. 

PRINCIPAL   AXES. 

157.  -At  every  jpoint  of  a  hody  there  are  at  least  three  ^prin- 
cipal  axes  perpendicular  to  each  other. 

When  three  axes  meeting  at  a  point  in  a  body  are  perpen- 
dicular to  each  other,  and  so  taken  that 

Xmxy  =  0,         Xmyz  =  0,         ^mzx  =  0  ; 
they  are  called  Principal  Axes. 


^'^^(^- 

B)  a)2&)3  =  L  ; 

^s'-^(^- 

-  C)  a)3&)i  —  M ; 

^'^^-(^- 

■  A  )  ft)i&)2  —  N. 

[157.] 


AXES. 


245 


The  planes  containing  the  principal  axes  are  called  Principal 
Planes. 

The  moments  of  inertia  in  reference  to  the  principal  axes  at 
any  point  are  called  the  Principal  Moments  of  Inertia  for  that 
point. 

Let  ONhe  any  line  drawn  through  the  origin,  making  angles 
a,  y8, 7,  with  the  respective  coordinate  axes. 
First  find  the  moment  of  inertia  about  the 
line  ON^.  From  any  point  of  the  line  (9iV, 
erect  a  perpendicular,  NP.  The  coordi- 
nates of  P  will  be  X,  y,  2.     Hence  we  have 

0N=  a?  cos  a  +  y  cos  yS  +  2  cos  7  ; 
1  =  cos^a  -}-  cos^iS  +  cosV 


Fio.  128. 


The  moment  of  inertia  of  the  l)ody  in  refercTice  to  ON^  will 
be 

7=  2m .  PiV2  =  2m  {OP'  -  OIP) 

=  2w  ■{  (x^  +  y^  +  z'^  —  (x  cos  a  +  ycoBfi  +  z cos 7)^  )- 

=  2m  ■{  {x^+p^  +  z'^)  (cos-o  +  co8-)3  +  003^7)— (a;  cos  a +y  cos  3+2  cos  7)*  )- 

=  2m  (2/2  4-  22)  cos'a  +  2m  (x'^  +  2-)  cos-/3  +  2m  {x-  +  y^)  CO8-7 

—  2l,myz  COS  fi  cos  7  —  22m2a!  cos  7  cos  a  —  22ma?y  cos  a  cos  /3 

= J^cos''a+jBcos')3+  ^008-7— 2^  cos  j8  cos  7— 2^  cos  7  cos  a— 2i^  cos  a  cos  j8  ; 

in  which  ^,i^,  6^, have  the  values  given  on  page  244,  and  D,  E^F^ 
are  written  for  the  corresponding  factors  of  the  preceding 
equation. 

This  may  be  illustrated  geometrically.  Conceive  a  radius 
vector,  r,  to  move  about  in  space  in  such  a  manner  that  for  all 
angles  a,  yS,  7,  corresponding  to  those  of  the  line  ON^  the  square 
of  the  length  shall  be  inversely  proportional  to  the  moment  of 
inertia  of  the  body.     Then 


1  = 


r" 


in  which  <?  is  a  constant.     Hence,  the  polar  equation  of  the 
locus  is 

-  =^  c>B'a+5co8'/8+C7co8'7— 22?  cos  3  cos  7— 2-^ cos  7  cos  a— 2i^co3  o  cos  j5. 


246  PRINCIPAL  :i58.j 

Multiplying  by  7^,  we  have 

which  is  the  equation  of  the  locus  referred  to  rectangular  coor- 
dinates, and  is  a  quadric.  Since  ^,^,  and  (7  are  essentially 
positive,  it  is  the  equation  of  an  ellipsoid,  and  is  called  the 
momental  eUijpsoid.  Therefore,  the  moment  of  inertia  about 
every  line  which  passes  through  any  point  of  a  body  may  be 
represented  by  the  radius  vector  of  a  certain  ellipsoid.  But 
_  every  ellipsoid  has  at  least  three  principal  diameters,  hence 
every  material  system  has,  at  every  point  of  it,  at  least  three 
principal  axes. 

If  the  ellipsoid  be  referred  to  its  principal  diameters  the 
coefficients  of  yz^  zx,  xy^  vanish,  and  the  equation  of  the  ellipsoid 
becomes 

c  =  Ax"  +  Bf  +  Cs",  .^^ 

■  \ 
In  many  cases  the  principal  diameters  may  be  determined  by 
inspection.  Thus,  in  a  sphere  ev6ry  diameter  is  a  principal 
axis.  In  an  ellipsoid  the  three  axes  are  principal  axes.  In  all 
surfaces  of  revolution,  the  axis  of  revolution  is  a  principal  axis, 
and  any  two  lines  perpendicular  to  each  other  and  to  the  axis 
of  revolution  are  the  other  two  principal  axes. 

^'  158.  Ifci,  hody  revolve  about  one  of  the  principal  axes  pass- 
ing through  the  centre  of  gravity  of  the  hody,  that  axis  will 
suffer  no  strain  froin  the  centrfagal  force. 

Let  s  be  a  principal  axis,  about  which  the  body  rotates.  The 
centrifugal  force  of  any  particle  will  be 


7?2a)^p  n:  mu?  Vx^-\-y^ ; 
which,  resolved  parallel  to  x  and  y,  gives 


mrn^x,         mco^y ; 


and  the  moments  of  these  forces  about  the  axis  of  z  are,  for  tlie 
whole  body, 

Jimay^xy,        Hmco^yx ; 


rl59.  160.1  AXES.  247 

but  these,  according  to  equations  (179),  are  zero.  If  the  body 
be  free  and  revolves  about  this  axis  it  will  continue  to  revolve 
about  it.  For  this  reason  it  is  called  an  axis  of  permanent  ro- 
tation. 

If  the  hocly  he  free,  and  the  initial  rotation  be  not  about  a 
principal  axis,  the  centrifugal  force  will  cause  the  instantane- 
ous axis  to  change  constantly,  and  it  will  never  rotate  about  the 
jpennanent  axis.  If,  therefore,  Ave  observe  that  a  free  body 
revolves  about  an  axis  for  a  short  time,  we  infer. that  it  revolved 
about  it  from  the  beginning  of  the  motion. 

RELATION  BETWEEN,  THE  AXES  .T,  7/,  C,  FIXED    IN    SPACE    AND   THE 
PRINCIPAL  AXES  a?i,  ^i,  ^i,  FIXED  IN  THE  BODY. 

159.  If  the  body  be  free,  take  the  origin  of  coordinates  (9, 
Fig.  129,  at  the  centre  of  the  mass ;  but  if  there  be  a  fixed 
point  about  which  rotation  is  forced  to  take  place,  take  the 
origin  O  at  that  point.  Conceive  a  sphere,  radius  unity,  hav- 
ing its  centre  also  at  0.  The  line  PP'  will  be  the  intersec- 
tion of  the  planes  xy  and  Xiyi^  and  P  one  of  the  points  where 
it  pierces  the  surface  of  the  sphere. 

Positive  angles  will  be  determined  in  substantially  the  same 
manner  as  positive  moments'*  described  in  Article  54,  page  80; 
thus,  positive  rotation  will  be  from  +  x  towards  4-  y,  from  +  y 
towards  +  2,  and  from  +  z  towards  +  x\  and  the  opposite 
directions  will  be  negative.  In  the  following  figure  the  rota- 
tions are  all  positive,  and  the  angles  or  amounts  of  rotation 
are  represented  as  less  than  90°. 

In  passing  from  one  system  of  rectangular  axes  to  another, 

*  It  may  be  observed  that  the  relations  of  the  axes  x,  y,  and  z,  in  the  follow- 
ing figures  are  not  the  same  as  on  the  preceding  pages.  Thus,  for  instance, 
on  pages  70  and  111  the  axis  of  y  is  vertically  upward,  while  in  the  following 
figures  z  will  be  in  that'position.  While  the  author  prefers  the  former  arrange- 
ment, for  the  reason  that  the  axes  x  and  y  would  then  retain  the  same  rela- 
tive position  in  the  printed  page  as  is  most  common  when  only  two  axes  are 
used  ;  yet  for  the  sake  of  conforming  with  the  usage  of  most  writers,  and  for 
greater  ease  in  comparing  results,  we  have  concluded  to  make  this  change. 
It  is  proper  to  observe  that  this  will  cause  no  change  whatever  in  the 
preceding  analysis,  provided  the  order  of  the  letters  be  observed,  to  the 
exclusion  of  righr&ud  left  handed  rotation. 


248 


RELATIONS   BETWEEN 


[159.] 


the  origin  being  the  same,  we  may  proceed  as  follows :  The 
system  may  be  turned  in  a  positive  direction  about  z  as  an 
axis,  bringing  OX  to  the  position  of  OP ;  then  rotating  it 
positively  about  OP  as  an  axis,  bringing  OZ  into  the  position 
OZx ;  and  finally  a  positive  rotation  of  the  system  about  OZ^^ 
bringing  OP  into  the  position  OX^  the  final  position  oi  OY 
being  OY^, 


Fig.  129. 

Then  let 

S  =  ZOZi,  being  the  angle  between  the  axes  ^  and  Si  which 
is  also  the  angle  between  the  planes  x^/  and  Xi  yi ; 

tp  =  XOP,  being  the  angle  between  the  axis  of  cc  and  the 
line  OP; 

cp  =  POXt,  being  the  angle  between  the  line  OP  and  the 
new  axis  of  x^. 

In  astronomical  language  the  line  OP  is  called  the  line  of 
nodes,  and  PXi  the  right  ascension. 

Expressing  the  above  steps  in  the  process  analytically,  'we 
have  Euler's  method  of  passing  from  one  system  of  axes  to  the 
other.  Thus,  let  x\  y\  z\  be  the  position  of  the  new  axes  at 
the  end  of  the  first  step,  then  we  have  {Coordinate  Geometry y 
Art.  54,  or  Art.  219)  * 


[159.] 


AXES. 


249 


x  —  x'  cos  xp  —  y'  sin  ^, 
y  =  x'  Binip  +  y'  cos  ^, 

2  =  Z  , 

Next,  turn  the  system  in  a  positive  direction  about  a?'  as  an 
axis,  through  an  angle  ^,  and  let  a?",  y\  z",  be  the  position  of 
the  axes  at  the  end  of  the  second  step,  then 

X   =  X  J 

y'  =  y"  cos  6  —  z"  sin  6, 

z'  =  y"  sin  6  +  2"  cos  ^. 

z 


Fio.  130. 


Fig,  131 


Finally,,  turn  the  system  in  a  positive   direction  about  z' 
through  an  angle  ^,  and  let  a?i,  3/1,  ^i,  be 
the  final  position  of  the  axes,  then 

x"  =  Xi  cos  (p  —  yi  sin  ^, 
y"  =  .Ti  sin  9?  +  ?/i  cos  <??, 

Z"  =  Za. 


Eliminating ,x*',  y',  z\  x\  y'\  2", be- 
tween these  equations  gives 

X  =  (cos  cp  cos  ^  —  sin  ^  sin  r\)  cos  G)  x^ 

+  (—  sin  cp  cos  ^  —  cos  9?  sin  \p  cos  ^)  yi 

+  sin  ^'  sin  ^  .  z-^ 
y  z=  (cos  ^  sin  ?/'  +  sin  <p  cos  tp  cos  ^  iTj 

-f  (—  sin  <;p  sin  ^  +  cos  (p  cos  ^  cos  0)  y^ 

—  cos  ^'  sin  B.  Zi 
s  =  sin  ^  sin  ^ .  a?i  +  cos  cp  sin  6^  .  y^ 

+  cos  ^  .  2i 


(181) 


260  RELATION  BETWEEN  [160.] 

We  have,  for  passing  from  a  system  of  rectangular  axes  to 
another  system  having  the  same  origin  {Coordinate  Geometry, 
Art.  219), 

X  =  Xi  cos  (xx^  +  ?/i  COS  {xy^  +  s^  cos  {xz^     \ 

y  =  a?i  cos  {ijx,)  +  y^  cos  {yy^  +  z^  cos  (y-j)      V  •         (182) 

z  —  x^  cos  ( zx^  +  ?/i  COS  (  2;yi)  +  %  cos  (^.a-i)      ) 

A  comparison  of  equations  (181)  and  (182)  gives: 

cos  {xx^  =  cos  cp  cos  ^  —  sin  «^  sin  tp  cos  0      \ 

cos  (,r?/i)  =  —  sin  ^  cos  ^  —  cos  ^  sin  ^  cos  ^  >-  .         (183) 

cos  (a?Zi)  =  sin  6^  sin  ^'  J 

cos  (ysci)  =  cos  ^  sin  ^j  4-  sin  <^  cos  tp  cos  <9 

cos  {yyi)  =  —  sm  cp  s,hi  tp  +  cos  9  cos  ip  cos  ^  J^  .         (184) 

cos  (y^i)  =  —  sin  6  cos  ^ 

cos  (zxi)  =  sin  ^  sin  6 

cos  (zyi)  =  cos  ^  sin  ^  y  .  (185) 

cos  ( zzi)  =  cos  ^ 

"We  here  liave  nine  direction  cosines  all  expressed  in  terms 
of  ^,  ip,  cp.  These  may  vary  while  the  axes  .Tj,  yi,  ^i,  remain 
lixed. 

The  same  relations  may  be  found  by  the  solution  of  spherical 
triangles  formed  by  arcs  of  great  circles  joining  the  points 
where  the  axes  pierce  the  surface  of  the  sphere  before  referred 
to. 

160.  Relation  hetween  the  angular  velocities  of  the  hody 
about  its  respective  principal  axes  fixed  in  the  body,  and  the 
angular  velocities  about  the  lines  OZ,  OZ^,   OP,  respectivel/y , 

In  Fig.  129,  ^-j-  will  be  the  angular  velocity  of  the  body 

about  OZ,  ^-j-  the  angular  velocity  about  (7Zi,  and  -j-,  the  an- 
gular velocity  about  OP.  For  the  ])urposc  of  representing 
these  quantities  in  the  simplest  manner,  and  for  greater  con- 
venience in  the  following  analysis,  let  the  value  of  -j-,  the  an- 


[160.] 


AXES. 


251 


gular  velocity  of  the  system  about  OZ,  be  represented  b}^  a 
definite  line  laid  off  from  (9,  positively  (in  this  case)  on  the 
line  OZ.  Similarly,  lay  off  a  line  of  proportionate  length  on 
the  line  OZ^  to  represent  the  angular  velocity  about  that  axis; 
and  similarly  on  the  line  OP,  This  representation  accords 
with  a  similar  representation  of  statical  moments  as  shown  in 
Article  171  of  the  author's  Elementally  Mechanics^  and  is  also 
in  accordance  with  the  remark  following  equation  (176)  of  this 
work.  In  a  similar  manner  oo^  will  be  represented  by  a  line 
on  aJi,  cwa  on  yj,  ojg  on  Zi.  This  being  done,  the  angular  veloc- 
ities may  be  referred  to  as  lines;  and  the  composition  and 
resolution  of  angular  velocities  be  treated  in  the  same  manner 
as  the  composition  and  resolution  of  forces.  Articles  55  and 
83.  Hence,  any  one  of  the  angular  velocities  will  equal  the 
sum  of  the  projections  on  that  axis  of  all  the  other  angular 
velocities.     Hence,  by  the  aid  of  equations  (185),  we  have  : 


cji  =  -j~  cos  ZOXi  +  -^  cos  Z^OXi 

4-  -17^^^  POX^  -f  &?2cos  YiOXi  +  gl>3  cos  ZiOXi 

=   -TT  COS  \ZX{)  +   -77-  COS  90    +  -77  COS  «> 

at  at  (If 

+  a?^  COS  90°  +  (i?3  COS  90° 

=  -^  sm  o?  Sin  ^  +   -7-  COS  m 
dt  ^  dt        ^ 

CO,  =  ~^  COS  ZO I\  +  '^  cos  Z,Or, 

-4-  -J-  cos  PO  Yi  +  0^1  cos  XiOFi  +  goq  cos  Z^O Ti 

dip  .     .      dd    . 

=  -r=7  cos  (D  Sin  a =-  sm  g? 

dt         ^  dt 

00^  =  -^  cos  ZOZi  +  -^  cos  Z^OZ^ 

+  T-  cos  POZi  +  (Wi  cos  XiOZi  +  00^  cos  YiOZ^ 
dtp         _      dcp 


(186) 


252 


SPONTANEOUS  AXIS. 


[161.] 


In  a  similar  manner  the  values  of  -rr^  -rrt  -rr  ?  ^^Y  be  found, 

dt^  dt^  dt^      -^  ' 

but  they  may  be  as  readily  found  by  elimination.     Eliminat- 
ing among  (186)  gives : 


dd 

Tt  ="^  ''"'  ^ 


GD^  sin  (p 


dtp        GOi  sin  (p  +  GJ2  cos  cp 

dt   "~  sin  6 

dm 


— -  ~  Qo^  —  cot  B  (qd^  sin  <p  +  G92  cos  cp) 


dt 


(187) 


AXIS  OF  SPONTANEOUS  ROTATION. 

161.  Considering  the  body  as  perfectly  free,  we  have,  ac- 
cording to  the  notation  immediately  following  equations 
(165):" 

fc  =  ^  +  iCi,  (188) 

from  which  we  readily  deduce  : 


dx       dx       dxi 


and  similarly, 


dt  - 

dt   "^  dt 

dy 
dt  " 

dy  dy^ 
dt   ^  dt 

dz 

dt  ~ 

dz  dz^ 
dt  ^    dt 

(188) 


X  being  a  coordinate  referred  to  a  system  of  coordinates  fixed 
in  space,  x  refers  to  the  centre  of  the  mass  in  the  same  sys- 
tem of  coordinates,  but  x^  is  a  coordinate  to  the  same  particle 
as  a?,  referred  to  a  parallel  system  having  its  origin  continually 
at  the  centre  of  the  mass.  Suppositions  may  be  made  arbi- 
trarily upon  any  one  of  the  three  terms  in  equations  (188),  and 
a  corresponding  relation  determined  between  the   remaining 


^^^. 


61 .]  SPONTANEOUS  AXIS.  253 


terms.     Thus,  if  -^  =  0,  and  similarly  for  ^  and  -^  ;  we 


have 


^=  $;  (189) 


which  shows  that  the  velocity  of  every  particle  is  the  same  as 
that  of  the  centre  of  the  mass.  Tliis  is  as  it  should  be,  since 
there  will  be  no  rotation. 

and  similarly  for  the  others ;  hence  the  velocit}^  of  any  parti- 
cle in  reference  to  the  fixed  origin  will  be  the  same  as  that  in 
reference  to  the  centre  of  the  mass.  This  is  evidently  as  it 
should  be,  since  the  centre  of  the  mass  will,  according  to  the 
hypothesis,  be  at  rest. 
Finally,  if 

dx       ^     dy       ^     dz       ^  HQIN 


re  have 

dt       ^'    dt  -^'    dt  -"' 

dx  _ 

dt  ~ 

dxi     dy            dyi     dz            dzi  ^ 
dt'    dt~  ~  dt'    dt~        dt' 

T*-)    7^  —  ~  "TtJJ         (1^2) 


Generalizing  equations  (191)  by  multiplying  by  ?,  m,  n^  re- 
spectively, and  integrating,  we  have 

Ix  —  a,        my  =h,        nz  =    c;  (193) 

where  «,  h,  c,  are  arbitrary  constants  of  integration.  Adding 
we  have 

Ix  +  my  =  a  -T-  h,     nz  -^  my  =  c  -i-  I;       (194) 

which  are  the  equations  to  a  right  line  in  space,  and  according 
to  (191),  this  line  will  be  at  rest  at  the  instant,  in  reference  to 
the  fixed  origin.  This  line  is  called  the  axis  of  sponta/neovs 
rotation. 


254  SPONTANEOUS  AXIS.  [1^2.] 

Since  iCi,  yi,  ^i,  equations  (192),  refer  to  tlie  same  particles,  it 
follows  from  these  equations  that  the  axial  velocities  of  the 
particles  on  the  axis  of  spontaneous  rotation,  in  reference 
to  the  centre  of  the  mass,  are  equal  but  directly  opposite  to 
the  axial  velocity  of  the  centre  of  the  mass  in  reference  to  the 
lixed  origin,  and  hence,  more  briefly  : 

The  velocity  of  the  axis  of  8pont(jmeous  rotation  in  reference 
to  the  centre  of  the  mass  is  equal  hut  in  ojpposite  direction  to 
that  of  the  centre  of  tJ^  mass  in  reference  to  a  fixed  origin, 

162.  To  illustrate  this  in  a  simple 
case,  let  AB  represent  a  body  whose 
i  centre  G  moves  in  the  line  CC  and 
3    which  rotates  about   its  centre.     If 
E  ED  represent  the  position   consec- 

utive to  AB,  they  will  intersect  in 
some  point  as  a,  then  will  a  line  through  a  perpendicular  to 
the  plane  of  the  two  positions  AB  and  ED,  be  the  axis  of 
spontaneous  rotation.  Conceive  the  line  AB  to  move  to  the 
consecutive  parallel  position  A'B',  GC  representing  the 
velocity  of  the  centre  ;  then  turn  the  line  about  G'  as  a  centre, 
that  point  a  which  has  the  same  velocity  backward  that  (7  had 
forward  will  be  a  point  in  the  axis  of  spontaneous  rotation. 
If  the  body  be  a  disc  moving  in  the  plane  of  the  paper  and 
having  a  uniform  rotation  about  its  centre,  the  spontaneous 
axis  will  have  a  uniform  motion  parallel  to  the  line  GG\ 
The  combined  motions  of  translation  of  the  entire  body  and 
of  rotation  about  the  centre  of  the  mass  may  be  considered  as 
a  simple  instantaneous  rotation  about  the  spontaneous  axis.  It 
will  also  be  observed  that  the  angular  velocity  of  the  body 
about  its  centre  will,  at  the  instant,  be  the  same  as  that  of  the 
centre  about  the  axis  of  spontaneous  rotation  (equations 
(193)). 

QuEKiES.  1.  Is   the   spontaneous   axis   always   within    the 
body  ? 

2.  In  Fig.  133,  when  the  bar  AB  falls  into  the  line  of  mo- 
tion GG\  where  will  the  axis  of  spontaneous  rotation  be  ? 

3.  If  the  centre  of  the  line  AB  has  a  uniform  velocity  in  a 
straight  line,  and  at  the  same  time  the  line  has  a  uniform 


[163.J 


SPONTANEOUS  AXIS. 


255 


angular  velocity  about  its  middle  point,  required  the  path  of 
the  extremity  B. 

4.  If  the  line  AB  has  a  uniform  velocity  of  20  feet  per 
second,  and  a  rotary  velocity  of  10  turns  per  second,  required 
the  distance  from  the  centre  to  the  axis  of  spontaneous 
rotation. 

5.  If  the  uniform  velocity  of  the  centre  of  a  disc  be  v  feet 
per  second,  and  the  uniform  angular  velocit}^  in  the  plane  of 
the  disc  and  of  motion  be  g?  ;  required  the  distance  of  the 
spontaneous  axis  from  the  centre  of  the  body. 

Let  X  be  the  required  distance ;  then 

XQi)  =  v;    ' 


X  = 


00 


If  r  be  the  radius  of  the  disc,  x  will  be  <,  =,  or  >  than  r, 
according  as  v  is  <,  =,  or  >  than  roj.  Only  one  line,  equa- 
tions (194),  fulfils  the  condition  of  a  spontaneous  axis. 

BELATIONS  BETWEEN  THE  SPONTANEOUS  AXIS  OF  ROTATION, 
THE  CENTBAL  AXIS  AND  THE  LINE  OF  ACTION  OF  THE  RESULTANT. 


163.  Observing  that  a?,  y,  s,  equations  (174),  are  coordinates 
in  reference  to  the  centre  of  a  free  body  as  an  origin  (the  sub- 
scripts having  been  dropped),  and  hence  are  the  same  as  iCi,  yi, 
2i,  in  equations  (192),  we  have,  by  substituting  the  values  of 
the  former  in  the  latter, 


dv 
dz 


(195) 


which  are  the  equations  to  the  axis  of  spontaneous  rotation^ 
the  origin  of  coordinates  being  at  the  centre  of  the  mass.  The 


256  RELATIONS   BETWEEN  THE  [163.] 

equations  to  the  axis  of  instantcmeous  rotation  or  central  axis, 
in  the  same  notation  are,  equations  (175),  (restoring  the  sub- 
scripts), 


GOyXi  =  GOxJ/i    J 


(196) 


A  comparison  of  equations  (195)  and  (196)  shows  that  these 
lines  are  parallel ;  hence, 

Tke  spontaneous  axis  of  rotation  is  parallel  to  the  instanta- 
neous {or  central)  axis. 

Letting  Y  be  the  velocity  of  the  centre  of  the  mass,  a,  h\  c, 
the  angles  between  the  line  of  motion  and  the  axes  a?,  y,  z,  re- 
spectively, we  liave 

dx       ^-.  dy        -rr       7     dz        ^^  ,,^^. 

^  =  Fcos(2,    -^  —  V cosh,    -jr  =  Fcos  c.  (197) 

Substituting  these  in  (195),  we  find  for  the  distance  be- 
tween the  central  axis  (196)  and  the  axis  of  spontaneous  rota- 
tion (195),  (Coordlnobte  Geometry,  Appendix  III.),  omitting 
numeral  subscripts, 


-   / {px-^r^y)  cos^  a  +  2  cja;  093  cos  a  cos  c  +  (oa\  +  col)  cos^  c  * 
\/   ^ ^ ^ .(198) 

Let  the  axis  of  y  be  the  axis  of  instantaneous  rotati6n,'^en 


^2/ 


will  csdx  =  0,  (i?a  =  0,  and  we  have,  dropping  the  subscript, 

M  =  — ,  (199) 

*  This  may  also  be  written 

V 


{gdI  +  Go'^)  cos^  a  + 

2go^ 

OOy 

COS  a  cosb  +  {GOy 

+  ^l) 

cos 'ft 

ool 

+ 

COy+Ool 

,* 

/(<-%  + 

-J) 

cos^  T)  + 

^OOy 

CD 

J  cos  6 cose  +  (fi)* 

■+■ 

<"! 

) cos*  e 

00l  +  <^'y    +    0^1 


[163.] 


SPONTANEOUS  AND  INSTANTANEOUS  AXES. 


257 


a  result  readily  deduced  from  Fig.  133,  since  ultimately  we 
would  have  CC  =  V  =  Ca.  go',  lience  6^a  =  ^i  =  F  -r-  go. 

If  the  anguhir  rotation  and  velocity  of  the  centre  both  be 
uniform,  Ai  (198)  will  be  constant ;  and  it  is  also  evident  that 
the  linear  and  rotary  velocities  may  vary  proportionately  in 
such  a  way  as  to  make  Aj  constant.  This  is  readily  seen  in  the 
more  simple  case  of  equation  (199),  and  illustrated  by  ex- 
amples 3  and  4,  p.  206. 

If  any  number  of  forces,  i^,  i^,  i^3,  etc.,  act  upon  a  body 
producing  both  a  translation  of  the  centre  of  the  mass  and  ro- 
tation about  that  centre,  they  will  be  equivalent  to  a  single 
force  Jd  applied  at  some  point  of  the  body ;  for  we  have  (Eqs. 
(85)^  (87),  and  (86)), 


22^  cos  a-  =  A"  =  7t  cos  a 
2Fcos  /3  =  r=  72  cosh 
21^  cos  y  =z  Z  =  It  cos  c 


R  =  vx^  +  r^  +  z- 


(200) 


(201) 


from  which  i?  becomes  known,  and  hence  a,  h,  c,  equations 
(200),  also  become  known.  To  find  the  line  of  action  of 
It;  since  the  moments  of  the  separate  forces  are  known, 
21^  cos  y  .  iji  —  2F  cos  /? .  iSi,  =  L  (say)  becomes  known, 
and  similarly  for  the  others ;  hence 


Zy,  -  1%  =  L 
Xzi  —  Zxi  =  M 
Yx,-~Xy,  =  N] 


(202) 


where  aJi,  yi,  ^j,  are  the  coordinates  of  the  line  of  action  of 
R  in  reference  to  the  centre  of  the  mass  as  an  origin  ;  hence 
(202)  are  the  equations  to  the  line  of  action  of  the  resultant. 
The  third  of  (202)  is  a  consequence  of  the  other  two.  Since 
any  point  in  the  line  of  action  of  a  force  may  be  taken  as  its 
point  of  application,  the  point  (ajj,  yi,  z^  may  be  considered  as 
17 


258  SPONTANEOUS  AND  [163.] 

the  point  of  application  of  the  resultant.    To  find  where  the  re- 
sultant pierces  the  plane  yz^  make  iTi  =  0  in  (202),  and  we  have, 

y,  =  -  ^,  z,  =  J;  (203) 

and  similarly  for  the  other  planes. 

If  the  forces  reduce  to  a  couple,  we  have  7?  =  0,  and  there 
will  be  no  motion  of  the  centre  produced  by  this  system  of 
forces,  although  the  centre  may  have  a  motion  due  to  initial 
conditions.  In  this  case  the  left  members  of  (202)  all  reduce 
to  zero,  and  the  point  of  application  will  be  found  to  be  infi- 
nitely distant.  The  right  members  for  statical  equilibrium 
would  also  be  zero,  but  if  Z,  M,  iV,  one  or  all,  have  finite 
values,  they  will  produce  rotation,  and  must  be  placed  equal 
to  the  left  members  of  (168). 

The  inclination,  0,  between  the  line  of  the  resultant  (202) 
and  the  central  axis  (196),  will  be  (Coordinate  Geometry,  Eq. 
(6),  Art.  199),  dropping  the  numerical  subscripts, 

cos  6  —      ^  (204) 

V(X^  +    Y'  +  Z^')  {CsOl    +   C4   +   Cy|  ). 

If  ^  =  0,  Equation  (201),  cos  ^  =  775  which  is  indetermi- 
nate. If  the  forces  are  parallel  to  the  axis  of  ,r,  then  Y  =  0, 
Z  =  0,  Go^  =  0,  and 

cos  (9  =  0 ; 

that  is,  the  action  line  of  the  resultant  will  be  perpendicular 
to  the  central  axis,  and  hence  also  perpendicular  to  the  axis  of 
spontaneous  rotation.  The  shortest  distance  between  the  ac- 
tion line  of  the  force,  equations  (202),  and  the  spontaneous 
axis,  equations  (195)  and  (197),  will  be  (Appendix  III.,  Coor- 
dinate  Geometry),  dropping  the  numerical  subscripts, 

f  M+--VGOsh)  fcD   --  J]  +  {-  GD  +  Veosa)  ( X-  — ^) 


[163.] 


iNSTA]srrANi:ous  axes. 


259 


As  a  special  case,  let  tliere  be  a  single  force  (or  forces  having 
a  single  resultant)  acting  parallel  to  the  axis 
of  X  and  at  such  a  point  as  to  produce  rotation 
about  the  axis  of  y  only,  and  let  this  be   a 
principal  axis,  then  we  have : 


C  =  90°,   OOg.  =  0,  ODy=  OO^  GD^  =  0  '^ 

and  (205)  becomes,  omitting  subscripts, 


B 

Fig.  134. 


,      M      V 

1  =  —  -h  —. 


and  (204)  gives 


12 


00 


(206) 


cos  ^  =  0  ;         .-.  ^  =  90°, 


hence  the  spontaneous  axis  will  be  perpendicular  to  the  action 
line  of  the  force,  will  lie  in  the  plane  yz,  and  be  at  a  distance 
from  the  action  line  equal  to  the  value  given  by  equation  (206). 
The  first  three  conditions  immediately  preceding  equation 
(206),  in  equations  (167)  give,  after  integrating  once,  R  being 
considered  constant. 


dx 

m 

V 

dy_ 

dt 

=  ^1 

d2 

di~ 

<^2 

Rt 


(206a) 


where  ???,  the  mass,  is  used  to  distinguish  it  from  Jf,  the  mo- 
ment, in  (206)  above.  The  velocities  i\  and  ^'g  are  constant, 
and  are  unaffected  by  the  force  or  impulse  R.  If  c\  and  c^  are 
not  zero,  F"will  not  be  the  actual  velocity  of  the  centre,  but 
will  be  the  velocity  produced  by  the  resultant  j^.  From  the 
first  of  (206a)  and  (199)  we  have 


7?       '^'^ 


idJixGO 


(207) 


260  SPOXTANEOUS  AXIS.  [168.] 

In  equations  (180)  we  will  have 

G^i  =  0,     Gj^  =  Oy    B  =  '2m  {a^  +  z^)  =  mlc^  ; 

and,  considering  the  moment  of  the  forces  as  constant,  we  have 
by  integrating 

^  =  Jf;  (208) 

and  hence  (206)  becomes 

In  Fig.  134,  h  being  the  point  of  action  of  the  force  in 
reference  to  the  body  AB,  and  a  the  projection  of  the  axis  of 
spontaneous  rotation,  then  I  =  db^  h^  =  Oa.  Let  Ob  =  /i^,  then 
from  (209)  we  have 

l  —  hi—  hi  =  Y  "^ 

r.hA=  h';  (210) 

and  since  ki  is  constant,  it  follows  that  in  the  plane  containing 
the  action  line  of  the  force  and  the  centre  of  the  mass,  the 
spontaneous  axis  and  point  of  a])i)lieation  of  the  force  are  con- 
vertible. In  other  words,  if  a  be  the  point  where  the  spon- 
taneous axis  pierces  the  principal  plane  xz  when  the  force  is 
applied  at  b  in  the  same  plane,  then  if  the  force  be  applied  at 
a  the  spontaneous  axis  will  pass  through  K  (See  also  Prob- 
lem 8,  page  209). 

Although,  in  establishing  equation  (206),  we  have  as- 
sumed a  constant  force  applied  at  some  point  in  one  of  the 
principal  planes  of  the  body,  yet  these  conditions  are -not 
always  practicable.  Examples  4  and  5,  page  206,  illustrate 
the  case  when  the  forces  are  constant  and  parallel.  Even  in 
these  cases  it  is  necessary  also  to  assume  that  the  body  has  n- 
initial  velocity,  or  if  it  has  it  must  be  the  same  as  might  havt.' 


[163.] 


MOMENTS   OF  MOMENTUM. 


261 


been  produced  by  tlie  action  of  the  force,  or  forces,  under  the 
conditions  above  imposed. 

But  all  these  equations  are  applicable  to  the  case  of  instan- 
taneous forces,  or,  in  other  words,  of  impact,  the  initial  veloci- 
ties either  being  zero  or  abstracted  from  the  above  conditions. 

The  equation  to  the  line  passing  through  the  origin  of 
coordinates — which  is  at  the  centre  of  the  mass — and  the 
point  where  tiie  action  line  pierces  the  plane  yz^  equations 
(203),  will  be  found  by  dividing  one  equation  of  (203)  by  the 
other,  and  is — restoring  the  subscripts — 


y'  =  -  M,^'- 


(211) 


Let  ff  be  the  angle  between  this  line  and  the  axis  of  spon- 
taneous rotation,  and  we  have  (195),  (211),  (See  Appendix  III. 
of  Coordinate  Geometry)^ 


cos  &  = 


OOs^Ml   —    Gify^JV^l 


Vo^xi 


+    GJy; 


+  oo4 


VMl  +  JVl 


(212) 


For  the  case  of  impact  this  may  be  reduced  to  a  form  which 
will  lead  to  an  interesting  general  result.  Since  this  condi- 
tion excludes  accelerating  forces,  X  =  0,  etc.,  in  (167)  and 
(168),  and  integrating  once,  we  have 


^dt  = 


dy 

dt 


C\M^-£  =  C,,M'^=  6i;      (213) 


dz 
dt 


K 


<Jz, 


dyi 


^'y^dt-'"'^  dt 


) 


L, 


„/        df/i  dxA        -_ 


(214) 


where  Z„  Mi,  iV,  are  the  moments  of  the  momentum,  as  shown 
by  the  left  uieuibers  of  the  equations.     Substituting  in  (214), 


262 


ROTATION. 


[163,] 


for  -^  etc.,  their  values  from  (174)  (since  (174)  are  equally 

true  for  the  origin  at  the  centre  of  the  mass),  and  making  the 
axes  principal  ones  by  the  conditions  of  equations  (179),  we 
finally  have 


"^"1  -  :^>/7.(2/5  +  ^?)  -  A 


GOy^     = 


COz.    = 


M, 


_  M, 

:2m  (zf  +  0^  ~   B 


N, 


m 


1  -^m(^  +2/f) 


(215) 


Substituting  in  (212)  the  values  of  gd^^^  gOz^,  coy^,  from  (215) 
gives 


cos  6' 


C 


-  1 


YVi. 


(216) 


If,  in  (216),  B  =  G,cos  6'  =  0;  that  is  if  two  of  the  prin- 
cipal moments  of  inertia  equal  each  other,  the  spontaneous 
axis  of  rotation  will  be  perpendicular  to  the  line  joining  the 
centre  of  the  mass  and  the  point  where  the  action  line  pierces 
the  plane  of  those  moment  axes. 

If  the  line  of  impact  be  parallel  to  the  axis  of  x,  Zi,  equa- 
tions (215),    will   be  zero,    which   gives  go^^  —  0,  and  since 

-^,  and-^  will  also  be  zero,  these  in  (195)  show  that  the  axis 

of  spontaneous  rotation  will  be  in  the  plane  yz.  If  now  B  = 
C^  the  axis  of  spontaneous  rotation  will  lie  in  the  plane  of  the 
moment  axes  of  B  and  C^  and  be  perpendicular  to  the  line 
drawn  from  the  centre  of  the  mass  normal  to  and  intersecting 
the  line  of  the  impulse  where  it  pierces  the  plane  of  these  axes. 
Further,  if  the  line  of  the  impulse  be  in  the  plane  a??/,  making 
any  angle  with  those  axes,  then  j!/i  =:  0,  and  iVi  =  0,  and  the 


[164.]  CENTRE  OF  PERCUSSION.  263 

denominator  becomes  infinite,  and  hence  6'  =  90°  ;  hence  the 
axis  of  spontaneous  rotation  will  be  perpendicular  to  the  plane 
xy^  or  parallel  to  the  axis  of  z ;  and  since  this  result  is  inde- 
pendent of  the  angle  which  the  line  of  the  impulse  makes  with 
the  axis  of  a?,  it  will  be  true  when  it  is  parallel  to  that  axis,  and 
in  the  plane  xij^  in  which  case  the  spontaneous  axis  will  lie  in 
the  plane  yz  and  still  be  parallel  to  the  axis  of  z. 

This  investigation  beginning  at  equation  (211)  may  be  still 
further  generalized  by  drawing  from  the  centre  of  the  mass 
a  line  normal  to  and  intersecting  the  line  of  the  impulse  and 
finding  the  angle  6"  between  this  line  and  the  axis  of  spon- 
taneous rotation,  but  the  results  so  found  will  be  but  little,  if 
any,  more  general  than  those  given  above. 

CENTRE     OF     PERCUSSION. 

164.  It  appears  from  Article  162  and  Fig.  133,  that  the 
point  a  in  the  axis  of  spontaneous  rotation  may  be  considered 
at  rest  at  an  instant,  and  hence  if  the  elements  on  that  axis 
were  held  rigidly  in  space  when  the  body  is  struck  at  h,  Fig. 
134:,  the  axis  would  suffer  no  shock.  Such  an  axis  generally 
exists,  as  shown  by  equation  (205),  or  more  simply  by  equa- 
tions (206),  (209)  and  (210).  Any  point  in  the  line  of  im- 
pact is  called  the  centre  of  percussion  in  reference  to  the  axis 
of  spontaneous  rotation,  and  the  centre  of  percussion  in  ref- 
erence to  any  axis  is  any  point  which  may  receive  a  blow 
without  imparting  a  shock  to  that  axis.  Equation  (210) 
enables  one  to  find  the  centre  of  percussion  in  reference 
to  any  assumed  axis. 

In  the  case  of  a  compound  pendulum  acted  upon  by  gravity 
only,  the  resultant  force  passes  through  the  centre  of  gravity 
of  the  oscillating  body,  and  the  centre  of  percussion  is  farther 
from  the  axis  of  suspension  than  is  the  centre  of  gravity.  In 
the  ballistic  pendulum,  the  blow  should  be  at  the  centre  of 
percussion,  if  possible,  to  avoid  shock  upon  the  axis  of  suspen- 
sion. 

Queries.  1.  Can  the  centre  of  percussion  be  at  the  centre  of  gravity  of  a 
body? 

2.  Can  the  axis  of  spontaneous  rotation  ever  pass  through  the  centre  of 
gravity  of  the  body  ? 


264 


CONSEEVATION. 


[165,  166.] 


CONSERVATION    OF   THE    MOTION    OF    THE    CENTRE    OF   THE   MASS. 

165.  Any  condition  that  will  render  the  second  members 
of  (167)  zero,  gives 


and 


df~^' 

df    "' 

di'-^' 

(217) 

integrated 

gives 

dx 

dt  -  ^" 

Ay  _   ri 

dt  -  ^" 

ds      „. 

(218) 

p=:C\t  +  B^ 

"   ) 

(219) 

z=C4+  B^_ 

when  (7i5  etc.,  J5i,  etc.,  are  constants  of  integration.     Eliminat- 
ing t  gives 


x-B^  _y-  B^ 


X  —  Bx      z  —  Bz. 


64 


6; 


6i 


L\ 


(220) 


which  are  the  equations  to  a  right  line. 

Equations  (218)  give  the  velocities  along  the  axes,  and  are 
constant.     If  C  be  the  resultant  velocity,  we  have 


C=  ^G^  +  6V  +  C^ 


(221) 


The  second  members  of  (167)  will  be  zero  when  no  forces 
are  acting  upon  the  system  ;  also  when  a  system  of  forces  all 
act  towards  the  centre  but  have  zero  for  a  resultant;  also 
when  the  forces  are  the  mutual  actions  between  the  parts  of 
the  system,  for  the  resultant  of  such  forces  is  zero;  hence 

When  a  hody  or  systetn  of  bodies  is  acted  ujpon  hy  any  sys- 
tem of  forces  having  zero  for  a  resultant,  the  motion  {if  any) 
of  the  centre  of  the  mass  will  he  rectiU'iiear  and  uniform,     " 

CONSERVATION  OF  THE  MOMENT  OF  THE  MOMENTUM. 

166.  Any  condition  which  renders  the  second  members 
of  (168)  equal  to  zero,  will  give,  omitting  the  subscripts. 


[166-]  OF  THE  MOMENT  OF  MOMENTUM.  266 

and  similarly  for  the  other  two  equations. 

Integrating  gives  the  three  equations  (214).  Let  the  ac- 
tual path  of  any  particle  be  projected  in  the  coordinate  planes^ 
and  consider  that  projection  which  is  on  the  plane  xy.  Let  r  be 
the  radius  vector  to  any  point  of  the  projected  patli,  d  the  angle 
between  a  and  /•,  then,  (Fig.  Ill), 

x  =  rco&d^        y  =  r  sin  6 : 
,\dx=  —r  sin  OdO  +  cos  ddr, 
dy  =  r  cos  Odd  +  sin  6dr ; 

which  substituted  in  the  third  of  (214)  gives 

21717^"^  =JSr,.  (222) 

But  -jr  =  CO  is  the  angular  velocity  of  the  particle  m  about 

the  axis  of  2,  tod  its  velocity  measured  in  a  circular  arc,  mraj  its 
circular  momentum,  and  r.mroo  the  moment  of  the  momen- 
tum, hence  '2m7^co  is  the  entire  moment  of  the  circular  momen- 
tum. But  this  is  the  same  as  the  moment  of  the  momentum 
in  the  direction  of  motion.  For  if  O  be  the  origin,  OA  =  /•, 
AOC  —  d8,  Aj)y  a  tangent.  Op  perpendicular  to  the  tangent 
from  the  origin,  then  will  the  actual  moment  of  the  momen- 
tum be 


o 


d^     ^ 

where  cZ.9  =  AC.     But 

AC  :  AB  :  :  OA  :  Oj?; 


/.  AC  =  ds  =   -TT-  r 


Op 


266  INVARIABLE  PLANE.  [167.] 

which  substituted  in  the  preceding  expression  gives 

and  taking  the  sum  of  all  the  particles  gives  the  left  member 
of  (222).  But  the  right  member,  JVi,  is  constant,  being  the 
constant  of  integration.  The  same  result  may  be  shown  for 
each  of  the  other  coordinate  planes;  hence  the  actual  mo- 
ment of  the  momentum  about  the  central  axis  will  be  constant 
and  equal  to 


ir=  Vli  4-  Jff  +  JSri  (223) 

Hence,,  lohen  a  hody  or  system  of  bodies  is  acted  ujjon  hy 
forces  directed,  toioards  the  centre  of  the  mass  /  or  hy  mutual 
actions  between  the  jparticles  ;  or  any  system  of  forces  in  which 
their  resultant  moment  is  zero,,  the  moment  of  the  momentum, 
is  constant^  and,  the  moment  of  the  momentum  projected  on  any 
plane  is  also  constant. 

Article  150  admits  of  the  same  generalization. 

THE   INVAEIABLE   PLANE. 

\f^'  '  167.  Consider  that  the  moment  of  the  momentum  on  the 
respective  coordinate  planes  is  represented  by  a  line  perpen- 
dicular to  the  plane  and  of  proportionate  length ;  then  will  K^ 
since  it  is  constant,  equation  (223),  be  normal  to  a  fixed  plane. 
If  the  axis  of  z  be  the  axis  of  this  plane,  L^  and  J/j  will  be 
zero,  and  iVj  =  K]  that  is,  the  moment  of  the  momentum  on 
the  plane  xy  will  be  a  maximum. 

Hence,  under  the  same  general  conditions  as  in  the  pre- 
ceding article,  there  is  a  fixed  plane  in  reference  to  which  the 
moment  of  the  momentum  is  a  maximum,  and  constant.  The 
plane  on  which  the  projections  of  the  moments  of  the  ipo- 
mentum  are  a  maximum  is  called  the  invariable  plane. 

In  the  solar  system,  knowing  the  positions  and  motions 
of  the  planets  at  any  time,  the  position  of  the  invariable  plane 
may  be  found. 


[168.]  ATTRACTIOlf.  267 

MUTUAL    ACTION    BETWEEN    PARTICLES. 

168.  The  mutual  action  between  particles,  or  bodies,  is  of 
the  nature  of  attraction,  or  of  repulsion,  or  of  both  these 
forces.  We  will  lirst  consider  attraction.  Newton's  law  of 
universal  gravitation  k — 

Two  particles  attract  each  other  with  a  stress  directly  jyropor- 
tioiial  to  th^  product  of  their  inasses,  and  inversely  as  the 
square  of  the  distance  hetwecn  them. 

Thus  if  m  and  m  are  two  masses  considered  concentrated 
at  mere  points — in  other  words,  the  masses  respectively  of  two 
particles — 11  the  stress  due  to  the  mutual  attraction  of  two 
units  of  mass,  the  one  upon  the  other,  at  distance  unity  be- 
tween them,  j?^  the  stress  between  the  masses  7H  and  iii  due  to 
their  mutual  attractions  when  the  distance  between  them  is  ic, 
then 

F=n^^,  (224) 

where  F^  m,  and  in!  are  in  the  same  nnits  as  77.  Taking  the 

origin  of  coordinates  on  the  line  joining  ^  ^  -^ 

the  particles,  af  the  abscissa  of  7)i\  x"  t^- 

of  711,  and  X  the  distance  between  them  at  ^       '^ 

time  t,  then  for  j?i  we  have  Pia.  i36. 


d'x 
m'  -.^  =  i^cos  180°  =  -  77 


7nin 


df     ~  "     ""^  """     -  ^^      9? 


dJ'x' 

_;/? 

or 

d^ 

=  -^^' 

and  for 

m. 

(Px" 

m 

df 

=  7^cos0° 

or 

d^-x" 

df 

(225) 


7nvi 


(226) 
I     also 

X  X      —   X, 

differentiating  which,  gives 


268  ATTRACTION  OF 

^x'  —  drx"  =  cPxy 
in  which  substituting  (225)  and  (226).  we  have 

—  =-n(m-^7n')—. 


[169.] 


dt^ 


Integrating, 


dr  ,  ,        ^  \   ax  J 


(227) 


(228) 


where   a  is  the   distance  between   the   particles  when  their 
velocity  is  zero.     Integrating  again, 


t 


^n  {m  4-  m') 


f(f^a;-iz^)l  +  «cos-'('|V   .    (229) 


In  the  preceding  investigation  the  mass  has  been  conceived 
to  be  concentrated  in  a  mere  point ;  it  will  now  be  extended  to 
that  of  a  mass  of  finite  dimensions. 

169.  To  find  the  attraction  of  a  homogeneous  spJiere  upon  a 
particle  exterior  to  it. 

Let  ABD  be  a  spherical  shell,  centre  (7,  P  tlie  position  of 

an  external  particle.  Conceive 
two  consecutive  radial  lines 
drawn  from  P^  cutting  the  shell 
in  the  points  A  and  J5,  and 
join  P  and  C. 

Let  B  =  APCy  y  =  perpen- 
dicular from  A  on  PC,  ds  =  an 
element  of  length  of  the  circle  at  A,  PA  =  r,  PC  =  c,  OA  ^ 
CB  =  p,  dp  =  thickness  of  the  shell,  d  =  density,  a  =  the 
radius  of  the  sphere,  and  7n  the  mass  of  the  particle  at  P. 

The  revolution  of  the  semicircle  about  PC  as  an  axis  \^ill 
generate  the  surface  of  the  shell.  The  area  of  a  section  of  the 
shell  at  A  whose  length  is  ds,  will  be 


Fig  13T 


dp .  ds. 


[169.]  SPHERES.  269 

Let  this  area  be  revolved  about  PC  as  an  axis  through  an 
angle  chp ;  it  will  generate  a  solid  whose  altitude  is  ydcp^  and 
the  volume  so  generated  will  be 

ydcp  ,dp.ds, 

and  its  mass  will  be 

Sydcpdpds, 

The  attractive  stress  between  this  element  and  the  element  at 
P  will,  equation  (224),  be  proportional  to 

m .  Sydcpdftds 

the  component  of  which  along  the  axis  PC  will  be 
«  d pdsy co%d  .  dcp 

and  the  attraction  between  the  entire  sphere  and  the  particle 
will  be 

^arrmdfr    ^^^^  (230) 

Jo  Jo  r^ 


In  order  to  integrate  this,  y,  6  and  r  must  be  found  in 
terms  of  ds  ;  or,  including  ds,  all  must  be  given  in  terms  of  a 
single  variable. 

Drop  the  perpendicular  C^:=p,  Fig.  137,  from  the  centre,  (7, 
upon  PB,  then 

jp  =  c  sin^; 
differentiating 

dp  =  ccosddd;  (231) 

also  from  the  triangle  PCA, 

r*  —  2rc  cos^  +  c^  =zp^; 


^0  ATTRACTION   OF  [169.] 

and  since  /3  will  be  constant  for  any  particular  shell,  we  have 
for  that  shell,  by  differentiating, 


dr 
d~d~  ' 

r  —  c  cos^ ' 

From  the  Theory  of  Curves, 

ds^  = 

dr'  +  T^d^, 

which,  combined  with  the  two  preceding 

equations,  gives 

ds 

pr 

dd  - 

r  —  c  cos^ ' 

and  this  with  (231)  gives 

c  cos^  ds 

fyrdp 

r* 

r 

From  the  figure  we  have 

r  —  c  cos^  =  Vp^  ~'!Pi 
also 

y  _p 

or 


r        G^ 


These  in  (230)  give 

4c7tmS  [^  ['    pdppdp  ^ 

C'        Jn    Jo    Vp'-f' 


0    •'0 


where  the  former  result  has  been  multiplied  by  2,  since  the 
same  perpendicular  CE  corresponds  to  two  elements  A  and  B, 
Integrating  gives 

and  integrating  again, 


SPHERES.  271 

— —J—  =  m  X  volume  of  the  sphere  x  -j 

=  m  X  ^^of^^Pf^^\  (232) 

That  is,  the  attraction  between  any  homogeneous  spliere 
and  a  particle  exterior  thereto  varies  directly  as  the  mass  of 
the  sphere,  and  inversely  as  the  square  of  the  distance  between 
the  particle  and  the  centre  of  the  sphere.  It  is  independent  of 
the  volume  of  the  sphere^  and  is  the  same  as  if  the  entire  mass 
he  considered  as  at  the  centre  of  tJie  sphere. 

For  the  same  reason,  if  the  mass  7??.  be  also  a  homogeneous 
sphere,  it  may  be  considered  as  concentrated  at  its  centre  of 
gravity ;  hence  the  attraction  between  any  two  spheres  varies 
as  the  product  of  their  masses  conjointly  and  inversely  as  the 
square  of  the  distance  between  their  centres.  This  result  is 
the  same  as  that  given  by  Newton  in  his  Prindpia,* 

'—  169a.  Mass  and  stress  are  often  referred  to  by  the  common 
name  pounds ;  but  when  necessary  they  are  distinguished  as 
pounds  of  mass,  or  pounds  of  force.  If  two  homogeneous 
spheres  of  equal  size  and  known  masses  be  placed  at  a  known 
distance  d,  from  each  other,  and  under  such  circumstances 
that  they  are  free  to  approach  each  other  under  their  mutual 
attractions ;  then  if  the  equal  opposite  forces  i^,  applied  to 
each  sphere  just  sufficient  to  prevent  their  approach,  be  ac- 
curately measured  in  pounds  by  an  accurate  balance,  we  have 
from  the  preceding  article  and  equation  (224) 

.•.iT=7^— ,.  (233) 

But  this  method  of  finding  Tl  is  impracticable,  chiefly  on  ac- 
count of  the  difficulty  of  measuring  the  exceedingly  small 

*  Principiay  B.  1,  Prop.  LXXVI.  Cor.  3. 


272  ATTRACTION  OF  SPHERES.  [169a.] 

value  of  n  which  would  result  from  manufactured  spheres  of 
manageable  size.  The  usual  method  of  finding  77  and  one  suffi- 
ciently accurate  in  practice,  is  to  consider  the  earth  as  a  liomo- 
geneous  sphere  whose  radius  is  the  mean  radius  of  the  earth. 
The  stress  due  to  tlie  attraction  between  the  earth  and  any- 
body at  its  surface  equals  the  weight  of  the  body,  or  m'  x  g 
=  -Fj  which  in  equation  (224)  gives 

n  =  ^,,  (234) 

where  J2  is  tlie  radius  of  the  earth  at  the  place  where  the  body 
is  weighed.  The  same  result  follows  from  equation  (227)  by 
neglecting  the  mass  m  of  the  small  body  when  compared  with 
that  of  the  earth,  and  substituting  g  for  the  acceleration,  and 
H  for  a?,  the  distance. 

Let  i?  =  20,850,000  feet,  the  mean  radius  of  the  earth; 

^  =  the  mass  of  the  earth,  being  about  5 J  times  that  of 

an  equal  volume  of  water,  page  227 ; 
g  =  32.16,  the  mean  value  of  the  acceleration  of  a  free 
body  on  the  earth,  p.  144  of  Elementary  Me- 
chanics ; 

then  from  (234)  we  have 


3  X  32|  X  2  _  67 

4  X  11  X  3.1416  X  20,850,000  ~  1,000,000,000 


lbs.  nearly. 


if  the  density  of  the  unit  sphere  be  the  average  density  of  the 
earth  and  a  foot  radius.  From  this  result  11  may  be  found  for 
any  assumed  material,  since  the  attractive  stress  will  vary 
directly  as  the  mass,  for  the  same  volume.  Equation  (227) 
thus  becomes  completely  determined.  The  unit  of  mass  for 
the  solar  system  is  sometimes  taken  as  that  of  the  earth  con- 
sidered as  5|-  times  that  of  an  equal  volume  of  water.  Equation 
(227)  is  for  motion  in  the  line  of  centres  of  the  bodies.     If 


[169b,  170.]  LEAST  ACTION.  273 

their  initial  motions  be  not  in  this  line,  the  bodies  will  describe 
orbits,  as  shown  on  pages  180  to  190.  In  the  solar  system  the 
mass  of  the  sun  so  far  exceeds  that  of  any  one — or  even  all — 
the  planets,  that  the  latter  are  neglected  in  determining  theoret- 
ically the  character  of  their  orbits.  The  value  of  TI  does  not 
enter  this  part  of  the  problem.  Having  proved  that  the  orbits 
are  ellipses,  their  magnitude  and  position  are  found  by  deter- 
mining three  points  from  observation. 

The  problem  of  'three  bodies'  subjected  to  mutual  attrac- 
tions and  having  arbitrary  initial  motions,  has  attracted  the 
attention  of  the  most  eminent  mathematicians.  In  the  case  of 
the  solar  system,  the  mass  of  the  sun  is  so  great  that  its  posi- 
tion is  considered  as  stationary,  and  the  problem  is  reduced 
chiefly  to  that  of  perturbations.  For  a  discussion  of  this  prob- 
lem see  works  on  Theoretical  Astronomy. 

169b.  Repiolsive  farces  of  the  nature  of  electricity  between 
two  bodies  are  supposed  to  vary  inversely  as  the  square  of  the 
distance  between  them. 

Elastic  forces  resist  the  displacement  of  particles  from  their 
normal  position,  and  vary  directly  as  the  amount  of  displace- 
ment, whether  it  be  a  resistance  to  tension  or  conipiession. 
In  this  case  it  may  be  found  that  the  orbit  of  a  particle  will  be 
an  ellipse. 

In  regard  to  constrained  motion  of  a  particle  on  a  curve  or 
a  surface  in  space,  equations  (167)  may  be  reduced  to  those  of 
(143).  If  the  body  be  finite  and  rotation  be  involved,  equa- 
tions (168)  will  also  be  necessary. 

All  the  equations  of  statics  are  also  contained  in  (167)  and 
(168)  by  considering  accelerations  as  zero.  The  resulting  for- 
mulas are  contained  in  the  preceding  pages. 

PRINCIPLE  OF  LEAST  ACTION. 

170.  Let  m  be  the  mass  of  a  particle,  v  its  velocity,  da  its 
path  during  the  time  dt^  then  is  the  value  of  the  definite  in- 
tegral 

m\    vda 
18 


274  LEAST  CONSTRAINT.  [170a.] 

defined  as  the  "  action  "  of  the  particle  in  passing  from  the 
former  position  Si,  to  the  latter  s^. 

Since  v  =  ds  -i-  dt,  the  above  expression  reduces  to 


pa 


^dt. 


The  principal  proposition  following  from  this  definition  is  : 
When  a  system  of  bodies  is  acted  u2Jon  hy  forces  directed 
towards  the  comrnion  centre  of  their  niass  ;  ,or  subject  only  to 
their  mutual  attractions  ;  or  hy  forces  tending  to  fixed  centres  ; 
then  in  Tnoving  from  a  given  jposition  to  another  the  sum  of  the 
actions  of  all  the  bodies  is  less  than  if  they  had  been  con- 
strained to  follow  any  path  diffei^ent  from  the  one  actually 
described. 

'  Tlie  proposition  in  this  form  is  due  to  Lagrange.  It  might, 
with  propriety,  have  been  called  ''the  principle  of  least  ms 
"mvaP  It  is  not  fruitful  in  the  solution  of  problems;  but  the 
proposition  once  proved,  leads  to  the  general  equations  (167) 
and  (168).  This  proposition  shows  that  if  a  particle  be  con- 
strained to  move  upon  a  surface  under  the  action  of  parallel 
forces  the  path  vrill  be  a  geodetic  curve. 


.o' 


GAUSS    THEOKEM  OF  LEAST  CONSTRAINT. 

170a.  If  a  system  of  onaterial  particles  be  in  motion, 
under  the  action  of  accelerating  forces^  the  sum  of  the  prod- 
ucts of  each  particle  and  the  square  of  the  distance  between 
its  place  at  the  end  of  time  dt,  and  the  place  which  it  would 
have  had  under  the  action  of  the  given  forces,  and  in  the  same 
initial  circumstances^  if  it  were  free,  is  a  minimum. 

This  theorem  of  Gauss  was  first  given  in  Crelle's  Journal, 
Yol.  I Y.,  1829.  It  contains  all  the  equations  of  motion  ;  and 
its  chief  interest  consists  in  presenting  the  subject  in  this 
peculiar  form. 


CHAPTEK  XII. 

MECHANICS  OF  FLUIDS.  I 

171,  Matter,  in  regard  to  its  physical  properties,  is 
infinitely  diversified.  The  physical  condition  is  conceived  to 
depend  upon  the  relation  between  the  attractive  and  repulsive 
forces  existing  between  the  particles  of  the  body.  Thus,  if 
the  attractive  forces  exceed  the  repulsive,  the  body  is  a  solid^ 
as  iron,  stone,  etc. ;  if  these  forces  are  equal,  the  substance  is  a 
liquid,  as  water,  alcohol,  etc. ;  and  if  the  repulsive  forces  ex- 
ceed the  attractive,  the  substance  is  called  gaseous,  as  air, 
hydrogen,  etc.  Solids  are  not  all  equally  rigid.  Thus,  steel 
is  vastly  more  rigid  than  jelly.  As  the  repulsive  forces  in- 
crease in  relation  to  the  attractive  ones,  the  bodies  become 
more  and  more  plastic,  as  glass,  iron,  lead,  jelly,  tar,  molasses, 
etc.,  passing  gradually,  and  it  may  be  imperceptibly,  from  the 
hardest  and  most  unyielding  substance  into  liquids.  If  the 
particles  of  a  liquid  are  not  free  to  move  among  themselves,  the 
substance  is  called  viscoits,  as  molasses,  vinegar,  etc.  Liquids 
also  pass  almost  irnpereeptihly  into  gases.  There  is  no  definite 
line  dividing  one  of  these  classifications  from  the  one  to  which 
it  is  more  nearly  allied,^and  since  it  is  impossible  to  reduce 
the  general  formulas  of  mechanics  so  as  to  make  them  practi- 
cally useful  for  all  the  cases  which  may  arise,  regardless  of  the 
properties  of  the  substance  considered,  we  make  arbitrary 
classifications,  as  indicated  above,  and  in  each  class  treat  of 
ideal  substances.  The  ideal  substances,  or  bodies,  are  perfect 
in  themselves,  and  may  not  have  a  single  representative  in 
nature  ;  still  the  results  deduced  on  these  hypotheses  may  rep- 
resent, with  a  more  or  less  close  approximation,  what  actually 
takes  place.  Tiie  more  nearly  the  real  conditions  approximate 
to  the  ideal  conditions,  the  more  nearly  will  the  equations 
represent  operations  or  facts  in  nature.     Discussions  involving 

275 


276  LAWS   OF  PRESSUEE.  [172,  173,  174.J 

the  imperfect  conditions  of  bodies — such  as  viscosity,  friction, 
elasticity,  etc. — ai'e  often  classed  under  Applied  Mechanics, 

172.  Definitions. — The  following  are  the  ti/incol  con- 
ditions usually  considered : 

A  KiGiD  BODY,  OY perfect  solid,  is  one  in  which  its  particles  are 
assumed  to  retain  their  relative  positions  under  tlie  action  of 
forces.     The  body  is  assumed  not  to  change  its  form. 

A  perfect  fluid  is  a  substance  in  which  its  particles  are 
perfectly  free  to  move  among  themselves.  The  property  of 
viscosity  is  thus  excluded. 

A  PERFECT  LIQUID  is  a  perfect  flicid  in  which  the  attractive 
and  repulsive  forces  are  equal.  Water  is  usually  taken  as  the 
type  of  such  a  substance. 

A  PERFECT  GAS  is  a  pcrfcct  fluid  in  which  the  repulsive 
forces  always  exceed  the  attractive  ones.  Such  a  substance 
would  expand  indefinitely  if  not  restrained.  Air  is  usually 
taken  as  a  type,  though  hydrogen  is  a  more  perfect  gas. 

A  heavy  fluid  h  one  in  which  its  weight  is  considered. 

173.  It  was  formerly  supposed  that  water  was  incompress- 
ible, while  it  was  known  that  air  could  be  easily  compressed, 
and  for  this  reason  fluids  were  divided  into  coinpressihle  and 
incompressiUe,  or  elastic  and  non-elastic  ;  the  former  of  which 
were  called  gases  and  the  latter  liquids.  Although  it  has  long- 
been  known  that  liquids  are  compressible,  yet  since  the  com- 
pVession  will  be  very  small  for  pressures  to  which  they  will 
ordinarily  be  subjected,  we  still  consider  a  perfect  liquid  as 
incompressible. 

LAWS    OF   PRESSURE. 

174.  The  pressure  upon  any  particle  of  a  perfect  fluid  at 
rest  is  equal  in  all  directions,  for  if  it  were  not,  there  would 
be  a  resultant  pressure  which  would  produce  motion  of  the 
particle,  since  it  is  assumed  to  be  perfectly  free  to  move. 

The  force  here  considered  is  finite.  The  weight  of  the 
particle  acted  upon  by  gravity  is  infinitesimal,  and  hence  if 


[175,  176.]  LAWS   OF  PJlESSURE.  277 

it  be  at  rest  the  upward  pressure  against  the  particle  must  ex- 
ceed the  downward  by  an  infinitesimal  amount,  the  amount 
being  equal  to  the  infinitesimal  weight  of  the  particle. 

175.  The  ipressure  of  a  perfect  fluid  nt  rest,  upon  the  sur- 
face of  the  vessel  containing/  ity  will  be  normal  to  that  surface 
at  every  point  of  it. 

For  otherwise  there  would  be  a  tangential  component,  and 
this  would  produce  motion,  which  is  contrary  to  the  hypothesis. 
This  proposition  is  also  true  in  regard  to  any  surface  exposed 
to  fluid  pressure  ;  hence,  if  a  body  be  immersed  in  a  fluid,  the 
pressure  upon  it  will  be  normal  at  every  point  of  its  surface. 
The  discussion  in  regard  to  the  stress  in  a  fluid  might  be 
founded  on  Article  D,  p.  154. 

176.  Eoery  external  pressure  upon  a  perfect  fluid  at  rest 
will  he  transmitted  with  equal  intensity  to  every  part  of  the  fluid 
in  the  vessel. 

For  if  there  were  any  unequal  pressure*  thus  transmitted, 
motion  would  result.  This  is  called  The  Law  of  Equal  Trans- 
mission. It  is  independent  of  the  form  of  the  vessel.  The 
pressures  due  to  the  weight  of  any  particle  of  the  fluid  will  also 
be  transmitted  equally  to  all  parts  of  the  fluid  in  the  vessel 
helow  the  point  where  the  paiticle  is  located.  It  cannot  be 
transmitted  above  that  point,  for  its  weight  is  equilibrated  by 
the  upward  pressure  at  that  point. 

This  proposition  is  illustrated  by  means  of  a  vessel  having 
closely  fitting  pistons  at  different  parts 
of  it,  and  the  vessel  filled  with  water, 
when  it  is  found  that  a  pressure  on 
any  one  of  the  pistons  [the  pressure 
being  p  pounds  per  square  inch] 
produces  the  same  pressure  per  square 
inch  on  each  of  the  other  pistons. 
The  pressure  per  unit  area  is  called 
the  intensity  of  the  pressure.  The 
proposition  is  not  rigidly  proved  by  i'lo-  i38. 

this  experiment,  for  the  pistons  cannot  work  without  friction. 


278 


LAWS   OF  PRESSUEE. 


[177,  178.] 


177.  The  pressure  upon  the  base  of  a  vertical  prismatic 
vessel  containing  a  perfect  heavy  fluid  equals  the  weight  of  the 
fluid ^te  the  pressure  upon  the  upper  surface  of  the  fluid. 

For,  according  to  the  preceding  article,  the  pressure  upon 
the  upper  base  is  transmitted  to  the  base  with  undiminished 
intensity ;  and  by  the  same  article  the  entire  pressure  due  to 
the  weights  of  the  particles  is  transmitted  to  the  base,  and  must 
then  be  supported  by  the  base. 

178.  The  pressure  upon  the  base  of  any  vessel  containing 
a  heavy  perfect  fluid  equals  the  weight  of  a  prism  of  the  fluid 
having  for  its  base  the  base  of  the  vessel,  and  for  its  altitude 
the  height  of  the  fluid ;  plus  the  pressure  per  unit  area  on  the 
upper  base  into  the  area  of  the  lower  base.  It  is  independent 
of  the  form  of  the  vessel. 

For,  if  over  any  element  of  the  base  a  vertical  prism  of  the 
fluid  be  conceived  of  the  same  weight  as 
that  of  the  liquid,  and  the  upper  surface 
be  subjected  to  a  pressure  of  the  same  in- 
tensity as  that  of  tlie  given  fluid,  the  press- 
ure on  the  element  will  equal  the  pressure 
on  its  upper  base  j;Z'W5  the  weight  of  the 
vertical  prism.  Article  177.  But  this 
pressure  will  be  transmitted  equally  in  all 
Fig.  139.  directions.    Article   176,  and  hence  pro- 

duces an  equal  pressure  on  every  element  of  the  base,  and  thus 
balance  the  real  pressures.  The  intensity  of  the  transmitted 
pressures  is  the  same  throughoui;  the  containing  vessel.  Article 
176.  If,  therefore,  the  vessel  be  oblique,  so  that  no  real  verti- 
cal prism  can  be  erected  on  any  element,  the  pressure  w^ill  re- 
main the  same,  depending  upon  the  vertical  height.  The  ideal 
vertical  prism  being  suppressed  the  proposition  will  be  estab- 
lished. 

\i  S  =  the  area  of  the  base  of  the  vessel, 

a  =  the  depth  of  the  fluid  in  tlie  vessel, 

d  =  the  weight  of  a  unit  of  volume  of  the  fluid, 

f  —  tlie  pressure  per  unit  of  area  upon  the  upper  base  of 

the  fluid, 
P  =  the  total  pressure  upon  the  base  of  the  vessel ; 


[179,  180.]  LAWS  OF  PRESSURE.  279 

then  we  have 

P  =  SaS+pS=  {Sa  +  p)S.  (236) 

179.  Static  Head. — In  the  preceding  Article,  conceive  a 
prismatic  vessel  liaving  the  same  base  and  filled  with  the  same 
fluid  to  such  a  height,  h,  as  to  produce  the  pressure  P  upon 
the  base,  then 

P  =  6hS,  (237) 

which  compared  with  equation  (236)  gives 

h=a+^,  (238) 

o 

This  value  of  h  is  called  the  head  due  to  the  pressure,  or  the 
reduced  liead.  It  is  a  height  of  the  fluid  which  would  produce 
the  actual  pressure  upon  the  base.  The  pressure  varies  directly 
as  the  head.  In  the  case  of  gases  confined  in  small  vessels,  the 
weight  of  the  fluid,  compared  with  the  external  pressures,  may 
generally  be  neglected,  in  which  case  we  have  (J  =  0,  and  (236) 
becomes 

P=:p8.  (239) 

Gases,  confined  in  a  vessel,  are  always  subjected  to  a  press- 
ure at  the  upper  surface ;  but  in  the  case  of  liquids,  p,  equa- 
tion (236),  may  be  zero,  in  which  case  a  becomes  the  same  as  h 
in  (237),  and  we  have 

^=M-i         (^*^> 

where  p'  is  the  pressure  upon  a  unit  of  area  of  the  base  of  the 
vessel. 

180.  Free  Surface. — If  the  upper  surface  of  a  fluid  is 
not  subjected  to  a  pressure,  it  is  called  2^ free  surface;  and  is 
sometimes  so  called  in  the  case  of  liquids  when  the  atmosphere 
causes  the  only  pressure.  Liquids  may  have  a  free  surface ; 
but,  according  to  the  definition,  a  gas  cannot  have  a  free  sur- 
face, since  it  may  expand  indefinitely.     When  the  surface  is 


280  MECHANICS  [181.] 

free  the  pressure  on  the  base  of  the  vessel  is  given  by  equa- 
tion (237). 

181.  Pressure  on  a  Submerged  Suefack —  The  sub- 
merged surface  may  be  the  interior  of  the  containing  vessel, 
or  the  surface  of  any  body  within  the  fluid,  and  of  any  form. 
The  normal  pressure  upon  any  element  of  the  surface  will  be, 
Articles  176  and  179, 

p  =  SxdA, 

where  x  is  the  reduced  head,  dA  the  element,  and  S  the  weight 
of  a  unit  of  volume  of  the  fluid  at  the  place  of  the  element. 
The  entire  normal  pressure  upon  the  surface  will  be 


P  =  J  dxdA,  (241) 

and  if  the  fluid  be  incompressible,  or  if  its  density  be  uniform 
throughout  the  surface  considered,  S  will  be  constant,  and  we 
have 


-!■ 


xdA.  (242) 


If  X  be  the  reduced  head  over  the  centre  of  gravity  of  the 
surface  considered,  then,  equations  (79),  taking  the  origin  in 
the  surface  of  the  fluid  vertically  over  the  centre  of  gravity 
of  the  surface,  we  have 


osA 


xdA; 


.     -  ,\P=d^A,  (243) 

that  is :  TTie  normal  pressure  of  a  fluid  against  any  sur- 
face submerged  in  it,  equals  the  weight  of  a  prism  of  the  fluid 
whose  hase  equals  the  area  pressed  and  whose  altitude  is  the 
reduced  head  over  the  centre  of  gravity  of  the  area  pressed. 
If  the  fluid  be  an  incompressible  liquid  having  a   free  sur- 


[182,  183.J  OF  FLUIDS.  281 

face,  the  reduced  liead  will  be  the  actual  head  over  the  centre 
of  gravity  of  the  area  pressed. 

182.  Resolved  Pressures. — The  pressure  in  any  fixed  di- 
rection will  be  the  sum  of  the  components  of  the  normal  press- 
ures in  that  direction.  If  6  be  tlie  angle  between  the  normal 
and  required  di^-ection  at  any  point  of  the  surface,  we  have 

resolved  pressure  =  6    xdA  cos  d, 

dA  cos  6  is  the  projection  of  the  element  on  a  piano  normal 
to  the  required  direction,  which  call  dB ;  then 

resolved  pressure  =  d  \xdB  =  6xB,  (244) 

where  x  is  the  reduced  head  above  the  centre  of  gravity  of  the 
projected  surface.  Hence,  when  the  projections  of  the  ele- 
ments are  not  superimposed 

The  component  of  pressure  of  a  hea/iyy  perfect  fluid  upon 
any  submerged  surface  in  any  direction^  equals  the  weight  of 
a  prism  of  the  fluid  having  a  hase  equal  to  the  projection  of 
the  surface  on  a  plane  normal  to  the  direction^  and  y)hose  alti- 
tude is  the  reduced  head  above  the  centre  of  gravity  of  the  pro- 
jected elements^  each  considered  to  be  at  the  depth  of  the  corre- 
ponding  surface  elements. 

183.  Resultant  Pressures. — If  a  body  be  submerged  in  a 
perfect,  heavy  fluid,  the  resultant  of  the  horizontal  pressures 
will  be  zero ;  for  the  projection  of  all  the  ele- 
ments upon  parallel  planes  will  be  equal,  and 
hence  the  opposing  pressures,  according  to  the 
preceding  article,  will  be  equal,  and  their  re- 
sultant zero.  For  the  same  reason  the  result- 
ant horizontal  pressures  upon  the  interior  sur- 
face of  such  a  vessel  will  also  be  zero. 

The  resultant  vertical  pressures  will  also  be  zqvOs  except  that 
due  to  the  weight  of  the  displaced  fluid.  Article  176,  hence 


a 

"f^^^: 

^m 

^^S 

282  MECHANICS  [184.J 

the  resultant  pressure  against  a  submerged  surface  will  be 
npward  and  equal  to  the  weight  of  the  displaced  fluid. 

184.  Centre  of  Pressure. — The  point  through  which 
the  resultant  of  all  the  pressures  upon  a  surface  passes,  is 
called  the  centre  of  pressure.  Let  AB  be 
a  submerged  surface,  CD  the  line  of  inter- 
section of  its  plane  with  the  free  surface, 
or  the  upper  surface  of  the  fluid  for  a  re- 
duced head.  Take  the  origin  at  any  point 
on  the  line  (77>,  y  along  ED,  and  x  along 
EF.  FG  =  the  head  on  the  element  cIA 
=  dxdy  at  F  Let  d  =  FEG  =  the  incli- 
nation of  the  surface  pressed  to  the  hori- 


FiG.  141. 

zontal,  then 


GF=z  x&in  6; 


and  the  pressure  on  the  element,  w  being  the  weight  of  a  unit 
of  volume,  will  be 

wdA  .  X  sin  6, 

and  its  moment  in  reference  to  the  line  CD  will  be 

w  dA .  ^  sin  /9, 

and  the  moment  for  the  entire  surface  will  be 

w  ^md  \  x^dA, 


' 


Denoting  the  distance  to  the  centre  of  gravity  of  the  area^ 
by  ^,  and  the  distance  to  the  centre  of  pressure  by  Z,  we 
have  for  the  reduced  head  over  the  centre  of  gravity : 

25  sin  6, 

and  hence  for  the  total  pressure  on  the  surface 

wActi  sin  ^, 


[184.]  OF  FLUIDS.  283 

and  for  the  moment  of  the  pressure, 

wA^  sin  6.1; 
hence  we  have,  by  equating  the  preceding  values, 

wAxl  sin^  =  w  sin^    ofdA; 

'■■■"'-^■.       (») 

that  is,  Articles  137  and  134,  The  centre  of  pressure  coincides 
with  the  centre  of  percussion^  the  axis  of  rotation  heing  i7i  the 
free  surface  ;  or  what  is  the  same,  it  is  the  moment  of  inertia 
of  the  surface  divided  by  its  statical  moment.  It  is  inde- 
pendent of  the  inclination  of  the  plane,  and  of  the  density  of 
the  fluid. 

Examples. 

1.  Required  the  entire  pressure  upon  the  interior  of  a  cone 
filled  with  xoater^  and  standing  on  its  base. 
Let  r  =  the  radius  of  the  base  of  the  cone,  and 

h  =  its  altitude. 
The  weight  of  a  cubic  foot  of  water  is  62J  pounds.     The 
area  of  the  base  will  be  tt?"^  ;  hence  the  pressure  upon  the  base 
will  be 

62i .  7z?\  h. 

The  normal  pressure  on  the  concave  part  will  be 


e2i.27rr.Wr^+  h^^h, 
or 

I  X  G2i  X  Ttrh  V/^  +  h\ 

which  added  to  the  preceding  gives 


284  MECHANICS  [184.] 

2.  Kequired  the  normal  pressure  upon  the  interior  of  the 
cone  in  the  preceding  example  when  inverted. 

3.  Eequired  the  normal  pressure  upon  the  interior  of  a 
spliere  tilled  with  water,  and  compare  the  result  with  the 
weight  of  the  water. 

4.  Find  the  normal  pressure  upon  the  interior  of  a  cylindric- 
al vessel  including  its  base,  when  filled  with  water. 

5.  Find  the  pressure  upon  the  interior  of  a  cone  filled  with 
water,  tlie  axis  being  horizontal ;  the  radius  of  the  base  being 
1  foot  and  the  altitude  4  feet. 

6.  Eequired  the  centre  of  pressure  of  a  plane  triancru- 
lar  surface  immersed  in  a  fluid,  the  base  being  in  the  free 
surface. 

The  moment  of  inertia  of  a  triangle  in  reference  to  its  base 
as  an  axis  is,  Article  104,  Example  3, 

Its  area  will  be  i  M^  and  the  distance  to  its  centre  of  gravity 
\  d ;  hence.  Equation  (245),  we  have 

7.  Eequired  the  centre  of  pressure  of  a  rectangle  hav- 
ing one  end  in  the  free  surface,  a  being  the  breadth  and  d  the 
depth. 

8.  Find  the  centre  of  pressure  of  a  rectangle  immersed  ver- 
tically in  a  fluid,  its  upper  end  being  a  distance  h  and  lower 
end  d  below  the  free  surface,  and  a  its  breadth. 

2  d^  -h^ 


Ans. 


3  d'-b^' 


9.  A  cone  standing  on  its  base  is  filled  with  water  ;  required 
the  vertical  pressure  upon  the  concave  part,  the  radius  of  the 
base  beino^  r  and  the  altitude  A. 


[185.]  OF  FLUIDS.  285 

10.  In  Example  9  show  that  the  pressure  upon  the  base 
minus  the  vertical  pressure  upon  the  concave  part  equals  the 
weight  of  the  water. 

11.  The  concave  surface  of  a  cylinder  filled  with  a  liquid  is 
divided  by  horizontal  sections  into  7i  annuli  in  such  a  manner 
that  the  pressure  upon  eachannulus  equals  the  pressure  on  the 
base  ;  the  radius  of  the  base  being  ;•,  required  the  altitude  and 
breadth  of  the  mill  annuhis. 

A71S.  Depth  A  =  7i7\ 
Breadth  of  mX\\  annxdus  =  \/rh  [V'^i  —^/m  —  1]. 

12.  A  rectangle,  breadth  14  feet,  depth  30  feet,  is  immersed 
vertically  in  a  liquid  with  one  end  in  the  free  surface  ;  I'e- 
quired  the  distance  below  the  free  surface  of  a  line  which 
divides  the  pressures  equally. 

Ans,  21.213  feet. 

13.  A  vessel,  in  the  form  of  a  paraboloid  of  revolution,  stand- 
ing on  its  base,  is  filled  with  water ;  required  the  normal  press- 
ure on  the  concave  part,  and  the  vertical  upward  pressure  on 
the  same,  the  radius  of  the  base  being  li  feet,  and  altitude  4 
feet. 

Flotation. 

185.  Consider  the  case  of  a  body  in  an  incompressible 
liquid. 

Let  Y  be  the  volume  of  the  body,  D  its  density  ;  V  the 
volume  of  the  liquid  displaced,  and  d  its  density.  Then,  ac- 
cording to  Article  183,  the  pressure  vertically  upward  will  be 

hence,  if  there  be  equilibrium,  we  have 

gDY=g8Y\ 
or 

Z__l.  (246) 


286 


MECHANICS 


[186.] 


that  is,  The  volume  of  the  hody  will  he  to  that  of  the  displaced 
liquid  as  the  density  of  the  liquid  is  to  that  of  the  hody. 
If 

then 


or  the  body  will  be  entirely  submerged,  but  if 

V>V'; 


then 


or  only  a  part  of  the  body  will  be  submerged,  and  the  body  is 
said  to  float. 

The  intersection  of  the  plane  of  the  free  surface  with  the 
floating  body  is  called  the  plane  of  flotation. 

The  line  joining  the  centre  of  gravity  of  the  solid,  G^  and 
the  centre  of  gravity  G  of  the  displaced  liquid  is  called  the 

axis  of  flotation^  and  if  this  line  be 
vertical  when  the  body  is  in  equi- 
librium it  is  also  called  the  line  of 
rest.  If  the  body  be  displaced  from 
its  line  of  rest,  the  vertical  through 
the  centre  of  gravity  C  of  the  dis- 
placed liquid  is  called  the  line  of 
support;  and  the  point  JIf  where 
this  line  intersects  the  line  of  rest  is  called  the  metacentre. 

For  the  equilibrium  of  a  floating  body  it  is  necessary  that 
the  line  of  support  shall  coincide  with  the  line  of  rest,  and 
the  equilibrium  will  be  stable  if  the  metacentre  for  an  indefi- 
nitely small  displacement  is  above  the  centre  of  gravity  of  the 
solid  ;  for  in  this  case  the  reaction  of  the  liquid  along  the  line 
of  support  tends  to  turn  the  body  toward  the  line  of  rest/  If 
the  centre  of  gravity  of  the  body  is  below  the  centre  of  gravity 
of  the  displaced  liquid,  there  will  also  be  equilibrium. 


Fig.  142. 


186.  The  depth  of  flotation  may  be  found  by  means  of 


[186.]  OF  FLUIDS.  287 

equation  (246)  when  the  density  and  form  of  the  body  and 
density  of  the  liquid  are  known. 

For  example,  to  find  the  depth  of  flotation  of  a  paraboloid 
of  revolution  with  the  vertex  downward  : 

Let  h  =  the  radius  of  the  base,  h  the  altitude,  and  x  =  the 
depth  of  immersion. 

Froui  the  equation  of  the  meridian  section  we  have 

also, 

"  ^  =  r^- 

The  volume  of  the  solid  will  be 

and  of  the  displaced  liquid, 

iTTj/x; 
and  these  substituted  in  Equation  (246)  give 


Examples. 

1.  Find  the  depth  of  flotation  of  a  solid  sphere  whose  radius 
is  6  inches  and  density  J  that  of  the  liquid  in  which  it  floats. 


MECHANICS  [187.] 

2.  Find  the  depth  of  flotation  of  a  cone  whose  altitude  is  5 
feet,  radius  of  the  base  8  indies,  and  whose  density  is  one-half 
that  of  the  liquid  in  wliich  it  floats,  the  axis  being  vertical  and 
apex  upward. 

3.  Required  the  depth  of  flotation  of  a  solid  paraboloid  of 
revolution,  base  downward,  radius  of  base  r,  altitude  A,  and 
density  §  that  of  the  liquid  in  which  it  floats. 

4.  Required  the  diameter  of  a  spherical  cavity  in  a  uniform 
spherical  shell  of  iron,  so  that  the  depth  of  flotation  shall  be 
equal  to  the  external  radius  of  the  shell,  the  external  radius 
being  ^•,  and  density  J  times  that  of  the  liquid  in  wdiich  it  is 
submerged. 

Ans,  r  = 


y  w 


5.  Required  the  pressure  necessary  to  just  submerge  a  cu- 
bical block  of  wood  each  of  whose  edges  is  a  feet,  and  whose 
density  is  i  that  of  the  liquid  in  which  it  is  submerged. 

6.  In  a  uniform  spherical  shell,  external  radius  r,  density  7, 
required  the  radius  of  the  cavity  that  the  plane  of  flotation 
shall  be  tangent  to  the  top  of  the  cavity. 

An^.  T  =  0.95/'-}-. 

SPECIFIC   GRAVITY. 

187.  If  an  external  pressure  act  upon  the  body,  either  forc- 
ing it  up  or  down,  thereby  producing  equilibrium,  we  have, 
w^hen  the  force  i^  acts  vertically  down  on  the  body, 

F+gDV  =  gdV';  (247) 

for,  the  weight  of  the  body,  gl)  V,  added  to  the  downward 
pressure  will  equal  the  vertically  upward  pressure  of  the  liquid. 
If  the  force  i^acts  upward,  then  the  upward  pressure  of  the 
liquid  and  the  force  i^  wall  equal  the  w^eight  of  the  body ; 
hence 

gDV=g6V'  -{-K  (248) 


[188,1 


OF  FLUIDS. 


289 


By  means  of  these  formulas,  the  weight  of  a  body  com- 
pared witli  the  weight  of  an  equal  volume  of  the  liquid  may 
be  determined.  If  the  liquid  used  be  selected  as  a  standard, 
the  relative  weight  thus  found  is  called  the  specific  weighty 
or  specific  gravity.  The  body  weighed  in  a  vacuum  gives 
directly, 

W=gDY\  ^'   (249) 

then  immersing  it  in  the  standard  liquid  and  ascertaining 
the  value  of  F  necessary  to  produce  equilibrium,  we  have 
from  the  preceding  equations, 


W=  gd  V  ±  F; 


or 


W 


gdV 


7=1± 


gdV 


(250) 


where  +  i^is  used  when  the  body  is  heavier  than  the  liquid, 
and  —  F  when  it  is  lighter. 

Water  at  a  fixed  temperature  (usually  60°  F.)  and  pressure 
(about  29.92  in.  of  the  barometer)  is  usually  taken  as  the 
standard.  For  a  further  development  of  the  subject  see  the 
Author's  Eleinentary  Mechanics. 

188.  Wlien  a  mass  of  liquid  is  in  motion  under  such  con- 
ditions that  its  form  becomes  permanent,  certain  problems 
pertaining  thereto  may  be  solved  by  the  principles  of  Statics. 
We  notice  the  two  following 


^F 


Pro  B  L  EM  s . 

1.  A  heavy  perfect  liquid  hamng  a  free  surface,  is  moved  in  a 
given  direction   with  a  con- 
stant acceleration,    required 
the  character  of  the  free  sur- 
face. 

Let  the  vessel  move  hori-  

zontally  under  the  action  of 
a  constant  force  F^  produc- 
ing an  acceleration/,  the  weight  of  the  liquid  being  W— the 
weight  of  the  vessel  being  neglected. 
19 


w 

Fio.  143. 


'290  MECHANICS  [188.] 

Then  we  have,  Equations  (20)  and  (21),  • 

W=Mg,    and    F=Mf\ 

.*.  -^r  =  -^  =  a  constant. 
^        / 

But  IP-i-  TT  is  the  tangent  of  the  angle  which  the  result- 
ant of  the  forces  of  the  free  surf  ace  makes  with  the  horizontal, 
which,  being  constant,  shows  that  the  slope  of  the  free  sur- 
face is  constant,  and  hence  it  is  a  plane. 

2,  A  free,  heavy,  perfect  liquid  in  a  cylindrical  vessel  is  ro- 
tated vnth  a  uniform  velocity  about  its  vertical  axis  ;  reqidred 

the  form  of  tliefree  surface. 

Since  the  forces  will  be  the  same 
in  every  meridian  plane,  we  may 
consider  the  form  in  one  meridian 
section,  as  xz  for  instance.  The 
acceleration  to  which  every  particle 
is  subjected  is  that  of  gravity, 
downwards,  and  the  resistance  to 
the  centrifugal  force  radially  in- 
ward. 

Let  00  be  the  constant  angular 
velocity;  then  will  the  centrifugal 
force  of  a  particle  at  a  distance  x 
from  the  axis  of  rotation  be  (Equa- 


FiG.  144. 


tion  (142)), 


also 

X  = 

—  moo^x, 

z  = 

-mg\ 

z 

9    , 
Go'x' 

which  is  the  tangent 

of  the  angle  of  the  resultant  force  with 

the  horizontal : 

dx 
"  dz 

_    9    . 

[189.]  OF   FLUIDS.  291 

which  integrated  gives, 

the  equation  of  the  common  parabola ;  hence  the  free  sur- 
face is^that  of  a  paraboloid  of  revolution  with  its  axis  ver- 
tical. 

Compare  this  result  with  that  of  Problem  8,  page  195. 

A  surface  to  wliicli  the  resultant  of  all  the  forces  at  each 
and  every  point  of  it  is  normal,  is  called  a  level  surface. 

Questions. — 1.  Will  the  parameters  of  all  the  paraboloids 
of  the  level  surfaces  at  different  depths  be  the  same  as  that 
of  the  free  surface  ? 

2.  Will  the  intensity  of  the  pressure  at  the  circumference  of 
the  base  of  the  vessel  be  the  same  as  at  the  centre  ? 

3.  If  the  revolution  be  so  great  as  to  cause  the  centre  of  the 
free  surface  to  touch  the  base  of  the  vessel,  or  even  to  expose 
a  portion  of  the  base,  will  the  free  surface  still  be  that  of  a 
paraboloid  of  revolution  ? 

Remark. — Since  the  paraboloidal  surface  is  produced  by  a 
uniform  rotation,  and  is  raised  from  a  free  horizontal  surface, 
it  will  be  subject  to  oscillations.  Could  steady  motion  be  con- 
tinued for  a  long  time,  these  oscillations  would  very  nearly 
disappear.  It  has  been  proposed  by  some  writers  to  resort  to 
this  principle  for  the  construction  of  very  large  concave  mir- 
rors for  astronomical  purposes,  but  the  delicate  physical  con- 
ditions, and  the  well-nigh  jy^7/(?c^  mechanism  necessary  for  its 
success,  are  obstacles  in  the  way  of  this  undertaking.  We  are 
not  aware  that  such  a  mode  of  making  a  mirror  has  been  at- 
tempted. 

FLUID    MOTION. 

189.  Definitions, — When  the  velocity  in  magnitude  and 
direction  at  every  point  of  a  fluid  vein  is  constant,  the  motion 


292  MECHANICS  [190.] 

is  said  to  be  "steady."  In  steady  motion  the  path  of  any 
particle  is  called  a  "  stream  line." 

If  throughout  a  finite  portion  of  a  fluid  mass  the  motion 
of  any  element  of  that  portion  consists  of  a  translation  and  a 
distortion  only,  the  motion  is  said  to  be  "  irrotational  " — a  term 
used  by  Thomson  and  others. 

When  the  particles  of  a  fluid  have  a  "  rotational "  or  "  vor- 
tex" motion,  a  line  drawn  from  point  to  point  so  that  its  di- 
rection is  everywhere  that  of  the  instantaneous  axis  of  rotation 
of  the  fluid,  is  called  a  "  vortex-line." 

If  through  every  point  of  a  small  closed  curve  we  draw  the 
corresponding  vortex-liue,  a  tube  will  be  obtained  called  a 
^*  vortex-tube."  The  fluid  contained  within  such  a  tube  con- 
stitutes a  "  vortex." 

190.  Bernoulli's  Theorem.— Let  a  particle,  in  steady  mo- 
tion, trace  the  stream  line  AB  ;  and  similarly  another  particle 

at  G  indefinitely  near  A,  trace  the 
stream  line  CD.  Trace  a  small 
closed  curve  through  the  points  A 
and  C\  then  will  all  the  stream  lines 
-  through  the  elements  of  the  curve 
F">- 145.  jiQ  form  the   elements   of  an  ideal 

tube,  and  may  be  replaced  by  an  actual  tube  conceived  to  be 
destitute  of  friction ;  and  since  there  is  steady  motion  the 
tube  will  be  filled  at  all  points,  and  constitute  an  element- 
ary stream.  The  quantity  of  fluid  passing  any  two  points, 
as  A  and  B,  in  the  same  time,  will  be  the  same,  and  in  this 
sense  the  flow  is  said  to  \>q permanent. 

Let  A^i  be  the  section  of  the  tube  at  J,  S  that  at  j5,  ^i  the 
velocity  of  the  flow  at  A,  v  that  at  B  ;  then 

SiVi  =  Sv, 

which  is  the  volume  of  the  liquid  flowing  m  a  unit  of  time. 

Letjr^i  be  the  intensity  of  the  pressure  exerted  at  A,  and  j? 
that  at  B,  then  will 

pA 


[190.]  OF  FLUIDS.  393 

be  the  entire  pressure  on  the  section  Siy  and 

will  be  the  work  done  hy  pi  in  a  unit  of  time,  since  the  veloc- 
ity may  be  considered  constant  for  an  element  of  time,  and 
may  represent  the  space  passed  over  in  a  unit  of  time.  Sim- 
ilarly, the  work  done  by^  in  the  opposite  direction  in  a  unit 
of  time  will  be 

—pSv, 

Take  any  datum  plane,  above  which  are  the  ordinates  Zi  to 
A  and  2  to  B,  then  the  work  done  by  gravity  while  the  liquid 
is  passed  from  the  height  2i  to  that  of  2,  will  be,  w  being  the 
weight  of  a  unit  of  volume, 

wSiVi{2i  —  z). 

The  difference  in  the  kinetic  energies  at  A  and  B  will  be 


2^ 


{v'  -  vl). 


Since,  according  to  the  assumed  condition,  there  is  no  re- 
sistance between  A  and  B,  the  liquid  between  these  points 
may  be  discarded — or,  what  is  better,  we  may  conceive  that 
the  liquid  just  before  and  behind  the  respective  elements  at 
A  and  B  serve  as  a  pair  of  perfectly  flexible  pistons  which 
yield  just  enough  to  keep  the  tube  full  at  all  points  passed  by 
this  elementary  mass.  Then  will  the  entire  work  done  upon 
this  mass  in  passing  from  A  to  B  equal  the  difference  in  the 
kinetic  energies  at  those  points,  or 

2hS,v,  ^pSv  +  wS,v,  (2,  -  2)  =  '^'  if-v^',         (251) 


or 


pA'^i  +  wSiv,2,  +  "^  v\  =pSv  +  wS^v,2  +  "^  v".  (252) 
zg  ^  '2g 


394  Bernoulli's  theorem.  [191.] 

In  this  equation,  piSiVi  is  the  potential  energy  of  the  ini- 
tial pressure  (Articles  26  and  151) ;  wS^ViZi  is  the  initial  po- 
tential energy  due  to  gravity  in  reference  to  any  arbitrarily 

ass^umed  horizontal  plane  ;  and  -^  ^  ^v^  is  the  initial  kinetic 

energy  of  the  mass  w  —  g.  The  sum  of  these  will  be  con- 
stant for  steady  motion.  The  second  member  of  tlie  equation 
represents  corresponding  quantities  for  any  point  of  the  stream  ; 
hence 

For  steady  unotion  without  resistances^  the  sum  of  tJie  poten- 
tial and  kinetic  energies  is  constant 

This  is  Bernoulli- s  Theorem.  It  may  be  expressed  in 
another  form,  for  dividing  Equation  (252)  through  by  wSiVi 
we  have 

•     Z.  +  ,.  + J  =  ^  +  .  +  f;  (253) 

w  2g      w  2g 

where  j9i  -r-  w  \^  the  head  due  to  the  initial  pressure,  z^  the  ini- 
tial head  in  reference  to  the  datum^  t^-^  2g  the  head  due  to 
the  initial  kinetic  energy  ;  and  similar  general  expressions 
apply  to  the  second  member. 

The  sum  of  the  heads  in  each  member  is  called  the  total 
head;  hence 

J^or  steady  7notion  without  resistances,  the  total  head  in  ref- 
erence to  any  Jurrizontal  plane  is  constant, 

191,  Discussion. — If  the  extremities  of  a  stream  in  steady 
motion  be  in  the  atmosphere,  the  pressure  at  the  ends,^i  and 
p,  will  be  that  of  the  atmosphere,  and  in  most  cases  will  be 
practically  equal  in  magnitude,  but  opposite  in  direction ;  for 
which  condition  (253)  becomes 

^  .  +  ^  =  ,  +  J?.  (254) 

'      2^  2^  ^     / 

The  initial  velocity  is  often  so  small  that  it  may  be  neglected, 
for  which  case  Vi  =  0,  and  (254)  becomes 

t7^  =  2g{z^-z),  (255) 


[192.] 


TORRICELLI'S  THEOREM. 


295 


and  if  the  datum  plane  passes  through  the  lower  end  of  the 
stream,  we  have  z  =  0,  and  (255)  becomes 


(256) 


wliich  re  called   TorricellPs   Theorem.     Comparing  (255)  or 
(256)  with  the  lirst  of  (16),  we  have 

In  steady  motion' without  resistances^  the  head  due  to  the 
velocity  equals  the  height  through  which  a  body  must  fall  to 
acquire  that  velocity. 

If  the  datum  plane  passes  through  the  lower  point  consid- 
ered, we  have  «  =  0,  and  (254)  gives 


V  =  Vv'i  +  2^01. 


(257) 


192.    To    REPRESENT   EQUATION  (253)    GRAPHICALLY,    let  AB 

be  a  stream  having  a  steady  motion  subject  to  different  press- 
ures along  its  path,  and  as-  £ 
sume  AHi  equal  to  Zi  and  ^^§0 
draw  the  horizontal  line 
HiH.  Let  AC  represent  the 
head  due  to  the  pressure  at 
A^  which  may  be  that  due  to 
the  atmosphere,  or  the  at- 
mosphere and  any  other  extraneous  pressure,  and  CE  the 
head  due  to  the  actual  velocity  at  A.  Then  will  AE  be  the 
total  head  above  A^  and  if  HiH  be  the  datum^  then  will  HiE 
be  the  total  head  above  the  datum. 
At  ^we  will  have 


Fig.  146, 


z  =  HB\        p-^w=^BF\ 


r8^ 


%j  =  BF, 


If  a  vertical  tube  be  inserted  in  the  stream,  having  an  open- 
ing up  stream,  the  liquid  should  rise  to  the  height  ED\  but 
if  it  be  turned  so  as  to  be  open  sidewise,  it  would  rise  only  to 
the  height  CF,     The  latter  is  called  the  hydraulic  head. 


296 


MECHANICS 


[193.] 


If  the  initial  velocity  be  neglected,  the  horizontal  line  ED 
will  pass  through  C. 

If  at  any  point  the  pressure  in  the  stream  is  zero,  the  line 
CF  will  be  depressed  and  touch  tlie  stream  at  that  point. 
Should  it  fall  below  the  stream,  the  pressure  would  be  nega- 
tive, but  as  liquids  have  no  tensile  strength,  this  condition 
would  destroy  the  "  steady  "  motion,  and  the  equations  would 
not  be  applicable. 

193.  If  a  vessel  of  varying  sections  be  left  free  to  discharge 
itself,  or  generally  if  a  fluid  has  a  "  steady  "  flow  through  a 
Na       pipe  of  varying  sections,  the  pressure  of  the 
fluid  in  the  small  sections  will  be  less  than 
that  due  to  the  statical  head,  frictional  re- 
sistances being  abstracted. 

Let  8i  be  the  section  at  D,  8  that  at  J?, 
then 

vS  =  ViSi ; 


.•.  v^  —vl  = 


SI  -  S'- 


'1 » 


Fig.  147. 

and  (253)  becomes 


w       w 


z  + 


S' 


2gS 


(258) 


where  z^  —  z  =  BD,  ^  =  head  due  to  the  pressure  on  D.  —  hx 

(say),  and  ^  the  head  due  to  the  pressure  at  B. 
If  8x  =  S^  w^e  have 


JL 
w 


h-V  DB\ 


(259) 


or  the  pressure  will  be  exactly  that  due  to  the  head. 

\i  S>  >6\,  the  last  term  of  (258)  will  be  positive,  and  hence 


[193.]  OF  FLUIDS.  297 

p-T-  w  will  exceed  the  head  given  by  (259),  and  we  may  write 
for  this  case 

^  =  Ai  +  £JV, 

w 

Let  the  section  at  A  be  less  than  bX  D  ov  S  <  S^^  then 

hence  AM  will  be  less  than  the  height  of  the  free  surface 
above  A,     Also 

^  =  Ai  +  AM, 
w 

If  S  is  so  much  less  than  ^Si  that 


is  negative,  then  p  -^  w  will  be  negative,  and  there  will  be  a 
tendency  to  a  vacuum.  Let  C  be  such  a  section.  Then  if  a 
bent  tube  CEG  be  inserted  at  C,  having  its  outer  end  below 
a  liquid,  the  fluid  from  F  will  rise  i'n  the  pipe  a  height  FGy 
so  that 

w 


FG  =  -B-  (260) 


Examples. 

1.  A  surface  elementary  stream  of  water  having  a  velocity 
of  16  ft.  per  sec.  undergoes  changes  in  its  sectional  area  as  it 
passes  a  vessel  which  are  proportional  to  the  numbers  4,  6,  4, 
3,  4,  6,  4 ;  in  what  way  can  the  head  remain  constant  ? 
Draw  a  vertical  contour  of  the  stream  with  figured  dimensions 
or  distances. 

We  have,  since  i^  varies  inversely  as  the  square  of  the  sec- 
tion, 

^  OC  ^-f,  -jjVj  iV>  i>  tV»  7V >  tV* 

f 


298  EXAMPLES.  [193.] 

Since  the  stream  is  in  the  surface^  will  be  constant,  being 
the  pressure  of  tlie  atmosphere,  therefore  z  or  the  height  must 
vary,  and  since  the  head  remains  absolutely  the  same  the  in- 

crease  of  z  must  be  the  same  as  the  decrease  of  t^-.   -y  =  16  on 

entrance,  and  -^  =  -—. — ^,  =  4  ft.    Hence,  as  the  successive 
2^      32  X  2 

values  of  ?r-  are 

|^  =  4,V-,4,V,4,V-,4, 
2  =  0,V,  0,-^,0,  V-,0; 

.__^  or  the  vertical  contour  will 
be  of  wave  form. 

Fig.  148. 

2.  "Water  flows  without  loss  of  head  through  a  horizontal 
pipe  of  a  diameter  varying  uniformly  from  3  in.  to  1  \h.  at 
smallest  section,  and  then  gradually  enlarges.  The  velocity 
on  entrance  being  T  feet  per  second,  what  will  be  the  mini- 
mum pressure  at  entrance,  in  order  that  the  pipe  may  run  full- 
and  what  may  be  the  maximum  diameter  of  exit  into  the  at- 
mosphere ? 

Since  the  pipe  is  horizontal,  z  =  Zi  in  Equation  (253),  and 
according  to  the  conditions  of  the  example,^  =  0  at  the  small- 
est section.     We  also  have 

V  at  entrance  =  7  feet ; 
/.  V  at  smallest  section  =  7  x  9  =  63 ; 
£i     49  _  {6S)\ 

.-.^1  =  26.44  lbs. 

For  maximum  diameter  of  exit  into  the  air  we  have  in  the 
same  equation  p^  =  0,  p  =  14.7  lbs.  per  square  inch,  Vi  =  63, 
andv  =  63  -i-d^; 


[193.] 


EXAMPLES. 


29^ 


14.7  X  144 

62.5 


C— 


2(/        2g 


.*.  d=  1.22  inclies. 

3.  Water  flows  from  a  tank  through  a  pipe,  the  lower  end 
of  whicli  is  15  feet  below  the  entrance,  the  sectional  area  of 
the  pipe  at  the  tank  end  being  twice  that  of  the  lower  end. 
Near  the  tank  the  pipe  is  perforated  or  broken;  lind  the  liead 
of  water  in  the  tank  necessary  to  prevent  air  leaking  in,  or 
water  out,  through  the  fracture. 

The  pipe  must  be  full,  and  the  motion  *i steady."  At  the 
ends  p=  pi  =  S4:  feet  (or  14.7  lbs). 

If  V  be  the  velocity  at  the  joint,  then  2v  will  be  that  at 
exit,  and  we  will  have 


34  +  1^  +  15  =  ^+34; 


2^ 


=  5  ft.  =  head  in  vessel. 


4.  Water  is  flowing  from  a  reservoir  through  a  siphon  pipe, 
the  discharge  end  of  whicli  is  20  feet  below  the  level  of  the 
reservoir.  The  diameter  of  the  pipe  is  2  inches  at  the  dis- 
charge end,  and  2|  inches  at  the  highest  point  of  the  siphon. 
Neglecting  all  resistances,  find  the  height  to  which  the  siphon 
may  be  raised  above  the  reservoir. 


At  A,  2h=0;   at  B,  j)  =  34  feet ;  and  Zi 
=  2  +  20  -f  A,  in  (253). 

Equating  beads  at  A  and  B, 


/^^ 


2<? 


Fio.  149. 


SOO  MECHANICS  [IW.] 

in  which  v  is  the  velocity  of  exit. 


■•— $i-(S)*! 


Equating  heads  at  0  and  ^, 


and 


^  +  34  =  20  +  34; 
.-.  v^  =  ^Og, 

A=f4+20{l-(gy}  =  25.7ft. 


If  6^^  =  /^i  we  would  find 

2yL^       \25 


=  34-/,+      [_i- 


and 

from  which  Ai  may  be  found  in  terms  of  A. 

FINITE    RESERVOIRS    AND   FINITE    ORIFICES. 

194.  If  the  liquid  in  a  vessel  of  finite  size,  Fig.  150,  is  free 
to  run  out  through  an  orifice  iu  the  base — or  side  of  the  vessel, 
the  course  which  the  elements  will  take  may  be  ob- 
served by  introducing  into  the  liquid  some  color- 
ing matter.  In  this  w^ay  it  is  found  that  the 
fillets  starting  from  the  upper  surface  form  curved 
paths  which  approach  the  orifice  as  a  common 


'I'll  i^i/iUW 


"q  point.  In  deflecting  the  paths  of  the  particles, 
Fig.  150.  centrifugal  forces  will  be  developed,  from  which 
it  follows  that  the  opposite  sides  of  the  fluid  streams  wilL  be 
subjected  to  unequal  pressures,  which,  however,  in  the  case  of 
perfect  fluids,  will  not  affect  the  flow,  except  as  it  changes  the 
length  of  the  path.  It  is  found,  also,  that  the  acceleration  of 
the  particles  is  different  in  the  different  elementary  streams. 


[104.] 


OF  FLUIDS. 


301 


As  the  streains  approach  each  otlier  at  the  orifice,  they  inter- 
fere, and  by  their  mutual  actions  produce  a  contraction  of 
tlie  vein  as  it  leaves  the  orifice  ;  the  point  of  greatest  contrac- 
tion being  at  a  distance  from  the  orifice  equal  to  about  one- 
half  of  its  diameter.  This  view  of  the  problem  of  flow  leads 
to  an  extremely  difficult,  if  not  strictly  impossible,  analytical 
solution.  We  therefore  adopt  a  more  simple,  and  at  the  same 
time  a  sufficiently  practical  hypothesis,  called  the  principle  of 
the  parallelism  of  sections^  which  implies  that  sections  parallel 
before  motion  remain  so  during  flow,  and  that  equal  volumes 
pass  tlie  parallel  sections  in  equal  times. 

If  A  be  the  height  of  the  free  surface  of  a  liquid  above  the 
section  of  greatest  contraction,  we  have  from  Equation  (255) 


7^  =  V^rjh  ; 


(261) 


or,  The  velocity  of  discharge  equals  that  of  a  hody  falling  in  a 
vacuum  from  the  free  surface  to  the  orifice. 

Experiments  show  that  in  some  cases  this  result  is  nearly 
realized  in  practice,  while  in  some  extreme  cases,  depending 
upon  the  form  and  conditions  of  the  orifice,  it  is  twice  too 
large.  In  practice,  the  several  cases  are  classified  as  mere  oi'i- 
fices,  short  tubes,  reentrant  tubes,  etc.,  etc.,  and  the  velocity 
as  determined  by  direct  experiment  in  each  of  the  cases,  divided 
by  the  theoretical  velocity,  is  called  the  coefficient  (or  niodulus) 
of  velocity. 

Thus  it  is  shown  that  when  the  discharge  is  through  a  thin 
plate,  or  past  a  well-defined  sharp  edge,  the  stream  is  at  first 
contracted,  forming  the  so-called  vena  con- 
tracta. 

The  diameter  of  the  section  of  greatest 
contraction  is  about  0.8  that  of  the  orifice ; 
hence  its  section  will  be  about  0.6-4  that  of 
the  orifice.  For  this  case,  the  actual  veloc- 
ity at  the  section  of  greatest  contraction  is 
found  to  be  about  0.97  of  the  theoretical,  and  hence  we  would 
have 

V  =  0.07  V'iyA"  (262) 


Fig.  151. 


802 


VELOCITY  OF  DISCHARGE. 


[195.] 


If  the  How  be  through  a  short  tube  wliose  sectioTi  is  the 
same  as  that  of  the  orifice,  it  is  found  that  the  quantity  dis- 
charged is  about  0.82  that  of  the  theoretical,  and  as  the  sec- 
tion of  the  stream  is  the  same  as  that  of  the  tube  the  entire 
loss  is  due  to  a  loss  of  velocity  ;  hence,  for  this  case 


V  =  0.82  V2g/i. 


(263) 


This  reduction  of  velocity  is  caused  by  the  interference  of 
the  fluid  veins  within  the  tube  near  and  about  the  section 
where  the  greatest  contraction  would  take  place.  If  a  small 
hole  be  made  in  the  pipe,  at  a  distance  from  the  inside  of  the 
vessel  equal  to  the  radius  of  the  pipe,  it  will  be  found  that  air 
will  rush  in,  thus  showing  that  there  is  a  negative  pressure 
in  the  pipe  at  that  point. 

Different  coefficients  are  found  for  other  cases. 
If  the  velocity  be  the  same  at  all  parts  of  an  orifice  whose 
section  is  k,  we  have  for  the  quantity  of  flow  in 
a  second — 

Through  an  orifice  in  a  thin  plate  : 


Fig.  152. 


q  =  0.64  X  0.97^  V^g/i  =  O.G2^  V2yA :   (264) 
and  through  a  short  tube : 

q  =  0.82Z;  V^gh.  (265) 


195.  If  the  orifice  is  so  large  as  to  cause  a  perceptible  ve- 
locity, 1?!,  of  the  free  surface,  we  have 

where  s  is  the  area  of  the  contracted  section  and  ^'^i  that  of  the 
free  surface.     This  value  of  v^  substituted  in  (254)  gives: . 


v  = 


(265a) 


[196.]  QUANTITY  OF  DISCHARGE.  308 

If  8  be  SO  small  compared  with  Si  that  it  may  be  neglected, 
we  reproduce  (261).  li  8  =  Si,  v  =  oo,  which  shows  that  this 
condition  cannot  be  realized.  The  liquid  would  drop  like  a 
free  body,  and  hence  its  velocity  would  not  be  dependent  upon 
liquid  pressure. 

Questions. — 1.  If  two  conical  vessels  of  the  same  dimensions  and  filled 
with  the  same  liquid  discharge  themselves  through  equal  orifices,  one  in  the 
base  and  the  .other  at  the  apex,  will  the  velocity  of  discharge  be  the  same 
when  the  heads  are  the  same  ? 

2.  In  the  preceding  question,  will  the  vessels  empty  themselves  in  the 
same  time  ? 

196.  If  the  orifice  be  in  the  side  of  the  vessel,  and  of  finite 
dimensions,  the  heads  of  the  several  ^lementary  streams  or 
fillets  will  be  different.  a 


Let  2  be  the  head  above  any  point  of  the  ori- 
fice, then  for  an   element  having   this   head,  we  b 
have                                                                                           c 


f 

where  h  is  the  coefficient  of  velocity ;  and  for  the  quantity 
flowing  through  this  element  in  one  second, 

q  =  c  V2(/2 .  dzdx^ 

where  c  is  the  coefficient  of  discharge ;  and  for  the  quantity 
flowing  through  the  entire  orifice,  we  have 


^=cV2^|jV2C?3</a?,  (266) 


integrated  between  such  limits  as  to  include  the  entire  area  of 
the  orifice.  The  free  surface  is  here  supposed  to  remain  at  a 
constant  height. 


E 


XAMPLE8 


1.  In  equation  (266)  let  the  orifice  he  a  rectangle^  h  the 
hreadth,  d  the  depth,  AB  =  h^,  AC  =  h^ ;  required  the  quan- 
tity which  woiddflow  through  the  orifice  in  a  unit  of  time. 


304  EXAMPLES.  [196.] 

We  have 

Q=  c  V2^\    J    Vz  dzdx  =  V2gl  c f  V^  c^2, 
•'/^i    0  J  hi  ' 

=  |c?V2^Z»(V-V)»  (266a) 

2.  If  the  upper  surface  of  the  rectangular  orifice  be  at  the 
free  surface,  the  opening  is  called  a  notch  ;  required  the  quan-- 
tity  discharged  through  the  notcli. 

3.  Determine  the  quantity  which  will  flow  from  a  triangu- 
lar aperture,  the  apex  being  in  the  free  surface,  h  being  the 
base,  h  the  altitude,  and  the  base  horizontal. 

•  Ans,  I  hhV2^. 

4.  In  the  preceding  example,  determine  the  flow  if  the  base 
be  in  the  free  surface.  Am.  ^hhV2gh. 

5.  Determine  the  quantity  of  flow,  if  the  orifice  be  a  circle 
whose  radius  is  7%  and  whose  centre  is  at  a  distance  h  >  r  be- 
low the  surface. 

Am.  nr^  VS^A  [l  -  i  g)'_  -^—  (£)\  etc.  ]. 

6.  To  deter7nine  the  time  hi  which  a  vessel  loill  einjyty  itself 
of  a  jperfect  liquid  through  an  orifice  in  its  hase. 

Take  the  origin  at  the  orifice,  z  vertical,  a  the  area  of  the 
orifice,  ^the  area  of  the  free  surface;  then 

Kdz 

will  be  the  elementary  volume  discharged  in  an  element  of 

time, 

aavdt  =  CLG  ^^gz  ,dt=  —  Kdz, 

the  quantity  passing  through  the  orifice  in  an  element  of'time, 
and  is  negative,  since  t  and  z  are  invei'se  functions,  hence 


L_  f^.ZL£^^  (266J) 

V2(J  Jo    Vz 


acV^g  Jo    V« 


[196.]  EXAMPLES.  305 

If  the  section  K  be  variable,  its  value  must  be  found  in 
terms  of  z  before  integrating. 

7.  To  find  the  time  in  which  a  prismatic  vessel  filled  with 
a  perfect  fluid  will  discharge  itself  through  a  mere  orifice,  a, 
in  its  base. 

2ZA 


Ans, 


8.  A  vessel,  formed  by  the  revolution  of  the  semi-cubical 
parabola,  })}/■  =  ^,  about  its  axis  2,  which  is  vertical,  is  filled 
with  a  liquid  to  the  height  h ;  to  find  the  time  in  which  the 
liquid  will  be  discharged  through  a  small  orifice,  section  «, 
at  the  vertex. 

Here 

K=nf  =  ^., 
and  the  limits  are  for  ^  =  0,  2  =  A,  and  f or  ^  =  ^,  «  =  0. 


Ana,  t 


9.  Find  the  time  in  which  a  paraboloid  of  revolution  whose 
altitude  is  h  and  parameter  ^,  full  of  liquid,  will  empty  itself 
through  a  small  orifice  at  its  vertex,  its  axis  being  vertical. 

Here  nij^  =  njpx, 

Ans.  t  = T==' 

dacV  2(/ 

10.  A  conical  vessel,  the  radius  of  whose  base  is  r,  and  alti- 
tude A,  is  filled  with  a  liquid  ;  required  the  time  in  which  the 
surface  of  the  liquid  will  descend  through  half  its  altitude,  the 
orifice  being  at  the  vertex,  and  the  axis  vertical. 

Here 

and  the  limits  are  z=-  h^  and  z  =  ih, 

^^^-  '  =      20acV-g     • 


306 


EXAMPLES. 


[196.] 


11.  Find  the  time  in  which  a  liquid  contained  in  a  parabo- 
loidal  vessel,  ^Z*  —  ^2,  will  descend  equal  distances  A,  the  flow 
being  through  a  small  orifice  whose  section  is  a,  at  its  vertex. 


Ans.  t 


caV2g 


The  times  are  equal  for  equal  heights  taken  anywhere 
along  its  axis.  This  would  foVm  a  water  clock  in  which  equal 
times  would  be  indicated  by  equal  spaces. 

12.  A  tank  whose  height  J^D  is  100  feet  above  the  level  of 
the  ground  is  supplied  by  a  1  inch  pipe   which  communicates 

with  a  IJ  inch  horizontal 
pipe  at  the  level  of  the 
ground,  and  is  fed  by  a  third 
pipe  2  inches  in  diameter 
proceeding  from  an  accumu- 
lator 3  feet  in  diameter,  with 
piston.  A,  10  feet  from  the 
ground,  loaded  with  30  tons. 
Neglecting  all  resistances, 
lind  velocity  with  which  the 
water  enters  the  tank. 


^ 


Fig.  154. 

Equating  heads  at  C  and  D  we  have 


30  X  2,240 


9  X  3.1416 


from  which 


X  62.5 


+  10  +  34=100  +^  +  34; 


V  =  64  feet  per  second. 


13.  Water  is  discharged  from  a 
vertical  rectangular  orifice  of  height 
d  and  area  A,     Show  that  approxi- 

mately  Q  =  c  V2gh  x  A{1  -g^^j, 

where  h  is  the  depth  of  centre  of  or- 

ifi(ie  below  the  water  level,  the  orifice  being  fully  immersed. 


Fig.  155. 


[197.]  DIVERGENT  TUBES.  307 

From  the  figure,  with  the  origin  at  the  upper  edge  of  the 
orifice  we  have 


or 

Developing  ( ^  +  ^  )  and  (^  —  q  )  to  four  terms,  and  per- 
forming the  operations  indicated,  we  have 

14.  Find  the  time  in  which  the  liquid  in  two  prismatic  ves- 
sels will  come  to  the  same  height,  one  discharging  itself  into 
the  other,  through  a  short  pipe  connecting  them  at  their  bases. 
Let  h  be  the  area  of  each  base,  k  their  height,  A  the  section 
of  the  pipe  ;  and  initially,  let  one  of  the  vessels  be  filled,  and 
the  other  empty. 

197,  If  a  conically  convergent  tube  BE  oi  the  form  of  the 
mna  contracta  be  attached  to  the  orifice  B^  and  to  the  small 
end  jE'a  tube  slightly  divergent  be  attached,  it  is  found  by  ex- 
periment that  the  amount  of  flow  is  in- 
creased, and  is  even  greater  than  if  the 
discharge  be  through  a  simple  orifice,  except 
when  the  flow  is  into  a  vacuum.  It  appears 
that  the  liquid  adheres  to  the  sides  of  the 
1^  156.  tube,  carrying  away  the  particles  of  air  from 

within  the  tube,  tending  to  make  a  partial  vacuum  at  E^  or  at 
least  to  diminish  the  internal  pressure,  thereby  making  more 


308  PRESSURE  OF  [197.] 

effectual  the  head  AB  and  the  pressure  of  the  air  on  the  upper 
surface. 

The  tube  being  entirely  filled,  it  is  the  case  of  steady  mo- 
tion, and  Bernoulli's  Theorem  applies. 

If  p  be  the  pressure  per  unit  of  the  atmospliere,  px  the 
pressure  at  E^  at  first  unknown,  and  considering  the  velocity 
of  the  surface  -^i  =  0,  equation  (253),  we  have  for  the  head  at 
A  (potential) 

W 

and  for  the  head  at  F^  z  being  zero, 


and  for  the  head  at  E^  z  also  zero, 


but  these  heads  are  all  equal,  hence 


A  +  £.=.^+^^J+£?;  (267) 

w      zg      w      2^      w 

.-.  ^  =  2gh\ 

whyjh  is  the  velocity  which  it  would  have  through  a  mere  ori- 
fice at  E\  but  as  the  section  at  i^is  larger  than  at  E,  the  flow 
has  been  increased,  and  hence  the  velocity  at  E  has  also  been 
increased.     This  becomes  apparent  from  the  equation 

v^Si  =  vS ; 

.-.i'.-l^,  (268) 

where  Vi  and  Si  apply  at  E,  and  S  SitE  which  is  larger  than  /Si. 


[197.]  FLUIDS   IN   MOTION. 

We  also  have  from  (267) 

|L  =  A  +  i^:^  =  7/, 

2^  w 

or  the  hydraulic  head  exceeds  h. 

To  lind  the  pressure  j^i,  we  have  from  (267)  and  (268) 


309 


(269) 


^,^^>  +  W.(l-|^); 


and  since  S-i  <  S,  we  have  jt>i  <  p. 
lipi  =  0,  we  have 


fS  y     p  +  wh 
\sj  ~      wh    ' 


(270) 


(271) 


and  if  the  liquid  be  water,  ^  -r-  t^j  =  34  feet,  nearly,  the  head 
due  to  the  pressure  of  the  air,  and  the  expression  becomes 


Since  pi  cannot  be  negative  for  steady  motion,  equation 
(270)  gives  the  limiting  ratio  of  the  sections,  but  this  limit 
cannot  be  quite  reached  in  practice. 

Eytelwein  found  that  when  the  mouthpiece  B^  was  shaped 
like  the  contracted  vein,  followed  by  a  divergent  tube  whose 
length  was  8{|  inches  long  and  angle  of  divergence  5°  9',  that 
2.5  times  as  much  water  was  discharged  as  through  a  simple 
orifice  of  the  size  of  the  section  at  ^,  and  1.9  as  much  as  through 
a  short  tube  of  the  same 
section  as  at  ^. 

If  two  vessels  be 
connected  by  a  tube 
as  shown,  and  filled 
to  the  same  heiglit 
with  the   same  liquid,  ^^o-  i56a. 

and  a  stream  be  established  in  any  manner,  it  will  continue 
to  flow  across  the  space  when  a  small  portion  of  the  tube  is 


310  REACTION  [198, 199.] 

removed,  provided  the  velocity  be  sufficiently  great,  and  will 
cease  only  when  overcome  by  friction,  or  by  the  difference  in 
heads  in  the  vessels. 

REACTION    OF    FLUIDS. 

198.  Newton's  Tliird  Law  of  Motion,  being  universal  in  its 
application,  includes  the  action  and  reaction  of  fluids.  If  a 
lieavy  fluid  discharges  itself  through  an  orifice  in 
the  side  of  a  vessel  suspended  by  a  cord,  the 
vessel  will  be  forced  away  from  the  stream  and 
the  cord  held  in  an  inclined  position.  The 
pressure  which  would  be  exerted  against  the  side 
of  the  vessel  if  the  orifice  be  closed,  is  removed 
when  the  orifice  is  open,  and  the  pressure  con- 
tinuing on  the  side  of  the  vessel  directly  opposite 
the   orifice   forces   the   vessel   in   that  direction. 


Fig.  1566.  There  is  also  a  pressure  exerted  in  the  same  direc- 
tion due  to  the  defiection  of  the  fluid  veins  from  their  course, 
as  will  be  shown  hereafter.     The  latter  is  called  a  reaction. 

199.  Centrifugal  Action. — The  force  wlr'ch  deflects  a 
body  from  a  tangent  to  a  curve  is  called  a  centripetal  force^ 
and  tlie  equal  opposite  action  of  the  body  upon  the  curve  is 
called  centrifugal  force.  Strictly  speaking,  we  should  say 
that  a  force  is  developed  between  the  body  and  the  curve, 
which,  acting  one  way  against  the  body,  forces  it  away  from 
the  curve,  and  in  the  opposite  direction  produces  an  equal 
pressure  against  the  curve.  If  the  body  be  attached  to  a  cen- 
tral point  by  means  of  a  cord,  the  centrifugal  action  would  be 
exerted  upon  the  fastenings  at  the  centre  ;  or,  if  there  be  no 
rigid  connection,  as  in  the  case  of  the  planets  moving  about 
the  sun,  the  centrifugal  force  would  be  the  same  as  if  the 
planet  moved  on  the  concave  surface  of  a  solid  coinciding  in 
curvature  with  the  orbit. 

If  the  motion  be  in  a  circular  arc,  let  m  be  the  mass  of  the 
body,  V  its  velocity,  r  the  radius  of  the  path,  and  (p  the  cen- 
trifugal force.  Since  the  centripetal  force  simply  changes  the 
direction  of  motion,  if  the  velocity  of  the  body  be  constant, 


[•>00.] 


OF   FLUIDS. 


311 


q)  will  also  be  constant.  At  A  the  body  will  be  moving  in  the 
direction  of  the  tangent  AB^  and  the  centripetal  force  will  act 
in  the  direction  J.  (9 ;  so  that  the  body  would 
reach  B  on  account  of  the  motion  in  the 
same  time  that  its  centripetal  force  would 
draw  it  to  2>,  the  points  B  and  D  being  con- 
secutive to  A, 


Let 

and 
then 


AOC=  6, 
AD  =  aj, 


a;  =  /•  (1  —  cos  S). 


Differentiating  twice,  dividing  by  d^^  and  multiplying  by  m 
gives 


m 


df 


mr  cos  6 


d^^ 


the  left  member  of  which  will  be  the  value  of  force  deflecting 
the  body  from  the  tangent  AB^  equation  (21). 


But  at  A, 

6>  =  0  and 

dd 
dt 

=   CD, 

the  angular 

velocity. 

Hence 

we  have 

9  = 

mroi^ 

=  m- 


^ 


(272) 


where  v  is  the  velocity  along  the  arc  of  the  circle. 

Equation  (272),  is  the  same  as  equation  (141),  and  also  the 
last  term  of  equation  (146). 


200.  Kesolved  peessubes. — If  a  particle  be  shot  into  a 


— X 


312  REACTION-  [200.] 

small  perfectly  smooth  tube  having  a  circular  bend  AEB^  it 

will  exert  a  uniform  radial  press- 
ure upon  the  tube  ;  the  compo- 
nents of  which  in  a  given  direc- 
tion as  CX  will  depend  upon  the 
position  of  the  particle.  The  sum 
of  these  components  for  a  given 
length  of  arc  will  be  the  same  as 
if  that  lengtli  were  full  of  such 
Fig.  158.  parti clcs  all  moving  with  the  same 

uniform  velocity  ;  and  if  such  a  stream  of  particles  be  contin- 
uous, the  pressure  will  be  constant. 

Suppose  then  that  a  steady  stream  of  fluid  passes  through 
the  tube,  and  let 

w  =  the  weight  of  a  unit  of  volume  of  the  fluid, 
7c  =  the  section  of  the  stream, 
T  =  the  radius  of  the  centre  line  of  the  tube, 
V  =  the  velocity  of  flow, 

s  =  AD  =  any  portion  of  the  path  from  the  initial  point  of 
the  curve, 

e=:AOD, 

CX  parallel  to  the  tangent  at  the  initial  point  of  the  curve, 
and  (7Z  normal  to  it. 

Friction  being  discarded,  the  velocity  will  be  uniform 
throughout  the  tube.  It  is  required  to  find  the  pressure  in 
the  directions  (7Xand  CY. 

For  any  element  of  length,  we  have  from  the  figure, 


and  from  (272) 
Integrating 


ds  =  rdd^ 


dcp='^Ms'-,  (273) 


w  =  — ks  — , 
^       g       r 


=  M?^,  (273a) 

r 


where  J!/" is  the  mass  flowing  into  the  tube  in  one  second. 


[200.]  OF  FLUIDS.  313 

Kesolving  dq),  equation  (273),  parallel  to  CX  and  CY  re- 
spectively, and  integrating  gives 

X,  =  -  M  f  sin  ^(i^  =  -  ^tr»  (1  -  cos  e\      (274) 

g       U  9 

ri  =  —  ki^  f  cos  Odd  =—kt^  sin  0,  (275) 

^         Jo  ^ 

If  the  angular  deviation  be  90°,  then   d  =  ^tt,   and  (274) 
and  (275)  become 


Xi,=-kv\  (276) 

r^,=--!cv>,  (277)' 

which  are  identical.     For  motion  through  a  semicircle  0  =  tt, 
and  we  have 

X„  =  2-kv^  (278) 

r,  =  0  ;  (279) 

the  last  of  which  shows  that  the  pressures  normal  to  the  line 

of  the  stream  balance  each  other.     For  the  entire  circum- 
ference 0  =  27t,  hence 


(280) 


We  observe  that  equations  (274)  to  (280)  are  independent 
of  the  radius  of  the  path  ;  hence  we  infer  that  the  path  may 
have  a  variable  radius,  but  cannot  be  zero,  since  equation 
(272)  will  be  infinite  for  r  =  0  and  v  finite. 

If  h  be  the  height  due  to  the  velocity  v,  then  v^  =  2ghy  and 
(276),  (277),  (278),  become  respectively 

Xi„  =  2wM,  (281) 

Ti„=2wkh;  (282) 

X,  =  ^M.  (283) 


314  REACTION  [201,  202.J 

Hence  : 

Deflecting  a  continuous  si/ream  of  frictionless  substance 
throug/i  an  a7igle  of  90°  along  a  cu7'vcd path,  produces  aj^ress- 
ure,  hoth  in  the  direction  of  the  initial  iriotion  and  normal 
thereto,  equal  to  a  prism  of  the  matter  whose  base  is  the  section 
of  the  stream,  and  whose  altitude  is  twice  the  height  due  to  the 
velocity.  Aiid  deflecting  such  a  streajn  180°,  the  head  due  to 
the  pressure  (283)  in  the  direction  of  the  initial  motion  will  be 
FOUR  times  the  head  due  to  the  velocity. 

If  JTbe  the  mass  of  the  liquid  flowing  through  any  section 
of  the  stream  in  a  unit  of  time,  then 


9 

(283a) 

and  equations  (274)  to  (278)  become 

X«  =  Jfo  (1  -  cos  0), 

(284) 

Ye  =  Mv  sin  d ; 

(285) 

X^=Mv, 

(286) 

Y,.  =  Mv; 

(287) 

X.  =  'i.Mv. 

(288) 

Equations  (284)  to  (288)  show  that  the  resultant  pressure 
due  to  deflecting  a  fluid  from  a  rectilinear  course  varies  first 
as  the  momentum  of  the  fluid  per  second,  and  second^  as  a 
function  of  the  angle  through  which  it  is  deviated. 

APPLICATIOISrS. 

201.  In  the  following  applications  all  resistances  due  to 
friction,  contractions,  enlargements,  or  whorls  and  eddies  in 
the  stream  will  be  discarded. 

202.  Discharge  from  the  side  of  a  vessel. — In  Fig. 
156^,  considering  that  the  fluid  filaments  have  their  origin  in 


[203,  204.J 


OP  FLUIDS. 


315 


the  free  surface,  their  initial  direction  will  be  vertically  down- 
ward, and  in  order  to  issue  from  the  orifice  horizontallj'  must 
be  deflected  through  an  angle  of  90° ;  hence,  equations  (281) 
and  (282),  the  pressure  on  the  opposite  side  of  the  vessel  due 
to  tJie  discharge  of  the  liquid,  will  equal  a  prism  of  the  liquid 
whose  base  is  that  of  the  contracted  section,  and  whose  height 
is  twice  the  head  above  the  orifice. 

Tlie  correctness  of  this  conclusion  in  regard  to  the  horizon- 
tal i)ressure  has  been  proved  by  one  Peter  Ewart,  an  English 
exi)erimenter,  who  determined  the  pressure  by  direct  measure- 
ment (Memoirs  of  the  Manchester  Phil,  *Soc.,  Vol.  IV.). 


203.  Discharge  vertically  upward. — In  this  case  the 
change  from  the  initial  direction  of  the  motion  will  be  180° 
and  equations    (283)  and  (279)  are  applicable,  if 
the  section  of  the  vessel  A  B  is  sensibly  the  same 
as  that  of  the  orifice  C,  in  which  case  the  pressure 
vertically  downward  will  equal  the  weight  of  a 
prism  of  the  liquid  wliose  base  is  the  section  of  the 
stream,  and  whose  altitude  is  FOUR  times  the  head 
due  to  the  velocity.    But  when  the  area  of  the  ori- 
fice is  small  compared  with  that  of  the  vessel,  the       fiq.  159. 
velocity  of  the  elements  in  A  B  will  be  small  com- 
pared with  those  in  the  immediate  vicinity  of  the  orifice,  and 
if  the  former  be  neglected,  the  downward  reaction   will  be 
nearly  that  due  to  the  deflection  through  a  quadrant  of  the 
elements  at  a  sensibly  uniform  velocity;  and,  hence,  equal  to 
a  prism  of  the  liquid  whose  base  is  the  contracted  section  of 
the   stream,  and   whose  altitude  is  twice  the 
head  due  to  the  velocity. 

204.  If  tlie  discharge  be  from  an  orifice  in 
the  base  of  the  vessel,  as  in  Fig.  IGO,  in  which 
the  orifice  is  so  small  compared  with  the  sec- 
tion of  the  vessel  that  the  velocity  of  the 
surface  may  be  neglected,  then  will  the  reac- 
Pio.  160.  tion  be  due  to  the  deflection  of  the  elements 


&^ 


316  KEACTION  [205,  206.] 

in  the  immediate  vicinity  of  the  orifice,  which  may  be  consid- 
ered as  of  uniform  velocity  and  nearly  all  deflected  through 
a  quadrant,  from  a  direction  nearly  horizontal  to  nearl}^  a 
vertical  direction ;  hence,  the  vertical  reaction  will  be  upward 
and  nearly 

Y  =  2wJch, 

or  the  upward  reaction  will  be  that  due  to  twice  the  head 
producing  the  velocity. 

205.  If  a  stream  of  liquid  impinge  normally  against  a  plane 
surface,  in  the  immediate  vicinity  of  the  intersection  of  the 

axis  of  the  stream  and  the  plane  an  eddy 
or  whorl  of  liquid  will  be  formed,  over 
which  the  stream  will  flow  as  along  a 
curve  and  be  discharged  tangentially  to 
the  plane  with  its  initial  velocity.  The 
direction  of  motion  being  changed  90°, 
Fig.  161.'"  equations  (281)  or  (286)  will  determine 

the  pressure  exerted  by  the  plane,  and  we  have 

r  =  2wkh  =  Mv;  (289) 

that  is  :  Thepressttre  exerted  hj  a  liquid  stream  flowing  nor- 
mally agai7ist  a  fixed  plane  equals  (numerically)  the  weight  of 
a  prism  of  the  liquid  whose  hase  is  the  section  of  the  stream^  and 
whose  altitude  equals  twice  the  head  due  to  the  velocity^  or  equals 
{numerically^  the  momentuvi  {per  secooid)  of  the  7nass  imping- 
ing against  the  surface. 

The  experiments  of  Miclielotti,  Weisbach,  and  others  show 
that  this  conclusion  is  very  nearly  realized  when  the  impinged 
surface  is  at  least  six  times  that  of  the  section  of  the  stream, 
and  placed  at  a  distance  of  not  less  than  twice  the  diameter  of 
the  stream  from  the  oriflce. 

206.  Cup  vane. — If  the  axis  of  a  stream  coincides  with  that 


207.J 


OF  FLUIDS. 


sn 


of  the  axis  of  revolution  of  a  surface,  and  impinges  against 
tlie  concave  surface,  they  will  be  deflected  as  before,  and  flow- 
ing along  the  eddy  as  along  a  curve,  the  flla-  q 
ments  will  leave  the  surface  tangentially,  and 
equation  (284)  will  be  applicable. 

If  the  tangents  to  the  surface  at  0  and  D 
are  parallel  to  the  axis  of  the  stream  AB,  we 
have  0=  7T,  and  (283)  or  (288)  will  be  applica- 
ble, and  we  have 

F  =  iwl'h  =  2Mv,  (290) 

The  resultant  pressxLve  due  to  the  impulse  of  a  liquid  stream 
against  a  concave  hemisphere  equals  tlie  weight  of  a  jprism  of 
water  whose  ha^e  is  the  cross  section  of  the  stream^  a/nd  whose 
height  is  four  iim£s  the  head  due  to  the  velocity. 

Weisbach  found  by  experiments  with  air  impinging  against 
a  concave  surface  that  the  pressure  was  about  0.S8  of  the  theo- 
retical value. 


207.  Bent  pipe. — If  the  tube  through  which  the  liquid  flows 
be  bent  through  an  angle  0  at  the 
point  B^  an  eddy,  or  whorl,  will  be 
formed  at  the  angle,  so  that  practi- 
cally the  flow  will  be  along  a  curve, 
and  equations  (284)  and  (285)  give 
the  resultant  pressures  parallel  and 
normal  to  the  initial  direction  of  the 
stream,  which  being  from  A  toward 
B  will  be 


Fig.  168. 


X  =  Mv{l-Qo^e) 
Y-Mv  sin  d 


(291) 


The  resultant  R  will  be 

B  =  VX^TY^  =  Mv a/2  (1  -  cos  0) ;         (292) 
which  alone  will  prevent  the  bodily  movement  of  the  tube 


318  EEACTION  [207.] 

were  no  other  external  forces  acting.  The  direction  of  R  will 
bisect  the  angle  ABG. 

In  practice  the  quantity  of  liquid  discharged  through  a  bent 
pipe  will  be  less  than  through  a  straight  one  of  the  same  sec- 
tion, on  account  of  the  contraction  of  the  stream  at  the  bend. 
The  mass  J[/'will  be  the  quantity  actually  discharged. 

If  two  forces,  each  equal  T^  were  acting  along  the  branches 
of  the  tube  and  away  from  the  angle,  and  of  sufficient  magni- 
tude to  produce  the  resultant  R^  we  would  have 

2^2  +  ^2  _2r2  (toQd  =  I^  =  2Jf  V  (1  -  cos  6), 

.\T=Mv\  (293) 

which  gives  the  required  value  of  the  force  T, 

If  there  be  two  bends,  B  and  (7,  in  the  pipe,  let  two  forces 

T  =Mv,  one  at  A  and  the  other  at  (7,  act  away  from  the 
angle  B\  they  will  hold  the  part  ABG. 
Similarly,  one  at  D  and  another  at  B^  each 
equal  to  T  and  acting  away  from  C  will  hold 
BCD\  but  the  equal  and  opposite  tensions 
along  BC  neutralize  each  other,  leaving  the 

^■4-^^^~|^^       tensions  at  A  and  D.      Hence,  if  frictional 
Fig.  164.  resistances  be  neglected, 

A  jperfectly  flexible  tube  of  umfoTTn  section  having  bends  of 
any  curvature,  and  its  ends  fixed  in  any  position^  will  not 
change  its  curvature  on  account  of  the  j^essure  due  to  the  flaw- 
ing of  a  fluid  through  it 

The  effect  of  the  weight  of  the  fluid,  which  is  not  included 
in  the  above  inference,  would  cause  the  tube  to  conform  with 
the  plane,  or  other  surface  on  which  it  rests. 

The  pressure  in  a  pipe  being  normal  to  the  curve  at  all 
points  will  be,  from  Equation  (c>),  page  139, 

T 

r        ^ ' 


[208,  209.]  OP  FLUIDS.  819 

but  from  (272)  this  becomes 

T= .  r  =  mv^  =  Mv. 

r 

as  before  shown  by  (293). 

208.  Impinged  surface  inclined. — If  the  plane  receiving 
the  impulse  be  inclined  an  angle  6  to  the  axis  of  the  stream, 
and  the  stream  be  confined  be- 
tween guide  plates  so  as  to  flow 
along  the  plane  in  the  line  of 
greatest  inclination,  the  case  will 
be  essentially  the  same  as  that  of 
a  bent  tube,  and  hence  the  press-  pio.  i65. 

ure  directly  opposed  to  the  stream,  and  also  normal  thereto, 
will  be  given  by  equations  (284)  and  (285)  respectively  ;  or 

P  =  Mv{l-  cos  e) ) 

209.  Remark. — Liquids  act  by  impulse  when  suddenly 
changed  in  direction  ;  and  we  have  seen  that  the  measure  of 
this  action  is  the  same  as  when  the  stream  flows  along  a  curve 
of  finite  radius.  In  practice,  however,  certain  resistances  fol- 
low an  impulse,  due  to  various  causes,  such  as  the  contraction 
of  the  stream,  eddies,  or  so-called  whorls,  which  make  the  effi- 
ciency of  the  fluid  less  than  when  it  acts  by  simple  pressure. 

Some  writers  improperly  use  the  term  impaci  in  this  con- 
nection, as  if  the  action  were  the  same  as  that  of  the  impact  of 
inelastic  bodies  ;  but  the  impact  between  liquids  and  solids  is 
only  infinitesimal  in  amount,  and  hence  eludes  measurement. 
The  true  action  is  not  an  impact  but  a  pressure — an  action  and 
a  reaction  of  finite  magnitude. 

The  stress  between  finite  solids  during  impact  is  rarely 
sought ;  and,  indeed  cannot  generally  be  found,  for  the  law  of 
action  is  generally  unknown.  To  find  it,  the  stress  as  a  func- 
tion of  the  time  must  be  known,  so  that  the  value  of  fl^dt 
may  be  found.     Also  the  law  of  the  distribution  of  the  stress 


320  EXAMPLES.  [309.] 

throughout  the  section  in  contact  must  be  known,  from  which 
it  appears  that  the  stress  may  be  variable,  and  the  finite  vahie 
souglit  will  be  the  sum  of  the  infinitesimal  stresses  acting 
upon  the  several  elements  of  the  section  of  contact.  Inelastic 
bodies  have  a  common  velocity  after  impact ;  and  if  a  small 
inelastic  body  impinge  against  an  indefinitely  large  one  at 
rest,  the  motion  is  destroyed ;  but  if  a  liquid  impinge  against 
such  a  body,  the  direction  of  motion  is  simply  changed. 

In  the  cases  above  considered,  both  the  momentum  and  the 
kinetic  energy  of  the  liquid  are  the  same  alter  the  impulse  as 
before. 

Examples. 

1.  The  nozzle  at  the  end  of  a  flexible  pipe  of  a  fire  engine  is 
directed  at  45°  to  the  horizon,  the  pipe  lying  along  the  ground. 
What  is  the  apparent  increase  of  weight  of  nozzle  when  150 
gallons  of  water  per  minute  are  discharged  with  a  velocity  of 
80  feet  per  second  ?     Also,  what  is  the  tension  of  the  pipe  ? 

For  the  tension  we  have 

T  =  Mv  =  AO  lbs. ; 
the  vertical  component  of  which  will  be 

r sin  45°  =  28  lbs.; 

which  is  the  apparent  increase  of  weight. 

2.  A  jet  of  water  of  sectional  area  A  impinges  beneath  a 
horizontal  plane  of  weight  TF".  Find  the  energy  of  the  jet  re- 
quired per  second  to  support  the  plane? 

Let  m  be  the  mass  of  a  unit  of  volume  of  water.  Then  the 
momentum  acquired  per  second  must  equal  TF, 


The  energy  of  the  jet  is 


,-Av'  =  W.  (1) 


^2 

wAv  X  — -,  (2) 

2g'  ^  ' 


[209.]  EXAMPLES.  321 

but  from  (1), 


^  ; 


and  substituting  in  (2), 


3.  A  vessel  containing  water  and  weighing  1  ton,  within 
which  the  pressure  is  9  atmospheies,  is  supported  by  discharg- 
ing water  downward ;  what  is  the  diameter  of  the  jet  \ 

Tlie  available  head,  supposing  the  discharge  against  the  at- 
mosphere, is  8  atmospheres  =  8  x  34  feet  of  water. 

The  momentum  of  the  jet,  and  the  consequent  reaction  is 


or 


—  ^v^  =  2,240; 

^'x5?:ix  2^  x8x  34  =  2,240; 

.'.  d  =  3.474  inches. 

4.  A  jet  of  water  strikes  a  fixed  vane  at  an  angle  of  30°  and 
then  glances  off  at  an  angle  of  60°  rela- 
tively to  the  tangent  through  the  initial 
point  of  impulse  ;  required  the  resultant 
pressure  on  the  surface,  the  jet  deliver- 
ing 600  gallons  per  minute,  with  *  veloc- 
ity of  10  feet  per  second. 


The  entire   deflection  of  the  jet  will 
be  90°;  hence 

H  =  V{Mvf  +  (Mvf  =  MvV^, 
600  X  8.4 


Fio.  100. 


60  X  32t 


10  X  \/2  =  36.9  lbs. 


322         ^  EXAMPLES.  [210.] 

5.  A  jet  of  water  4  inches  in  diameter,  velocity  25  feet  per 
second,  impinges  on  a  fixed  cone  at  its  vertex,  their  axes  coin- 
ciding and  the  apex  angle  being  30° ;  find  the  pressure  tending 
to  move  the  cone. 

P  =  —  Qv{l-cosla)=:  2—  Qv  sin^  7°  30'. 

=  3. 6  lbs. 

6.  A  jet  of  water  4  inches  wide  and  1  inch  thick  impinges 
tangentially  on  a  concave  cylindrical  surface  of  6  inches  ra- 
dius, flowing  over  it  in  a  stream  of  the  same  section  and  finally 
leaving  it  tangentially  after  being  deflected  through  an  angle 
of  60° ;  the  velocity  being  10  feet  per  second,  what  will  be 
the  intensity  of  the  normal  pressure,  and  the  resultant  force 
in  the  direction  of  the  jet  ? 

According  to  equation  (273<x)  we  have  for  the  entire  cen- 
trifugal force 

where  s  is  the  length  of  the  arc  of  contact,  and  dividing  by 
the  area  of  the  concave  surface,  which  is  -j*^  s,  gives  for  the  in- 
tensity of  the  pressure, 

r  ' 
or 

o       624       4        1       -^     10       on  ^  A 

^  ^  32f  ^  12  ^  12-^  ^^  ^T  "  "^ 

The  resultant  pressure  in  the  direction  of  the  jet  will  be 

Mv  (1  -  cos  60°)  =  2.7  pounds. 

210.  If  the  surface  impinged  upon  be  in  motion  and 
moves  in  the  initial  direction  of  the  stream  with  a  uniform 
velocity  u,  the  relative  velocity  will  he  v  —  u.  If  M  be  the 
mass  deflected  by  the  surface  per  second,  the  pressure  will  be 


[211-]  REACTION  OF   FLUIDS.  323 

the  same  as  that  of  a  stream  moving  with  the  velocity  v  —  u 
against  a  surface  at  rest ;  hence  if  6  be  the  deflection  of  the 
stream  relatively  to  the  surface^  we  have  only  to  write  v  —  u 
in  equations  (284)  and  (285)  to  make  them  applicable  to  this 
case,  observing  that  if  the  surface  moves  against  the  stream  u 
will  be  negative.     Hence,  we  have 

X9  =  M(v-  u)  (1  -  cos  6)  =  2M{v  -  u)  sin^  id,  (295) 

Yb  =  M(v  -  u)  sin  e ;  (296) 

r.A\„  =  M{v^u),  (297) 

Ti^  =  M{v^9i);  (298) 

and  similarly  for  other  values  of  0. 

WORK  DONE. 

211.  When  the  body — called  a  vane — receiving  the  impulse 
of  the  stream,  moves  in  the  direction  of  the  stream  or  at  an 
acute  angle  therewith,  work  is  done  and  energy  is  imparted  to 
it  and  the  mechanism  attaclied  thereto  ;  but  if  the  motion  be 
in  the  opposite  direction,  energy  will  be  imparted  to  the  fluid. 

If  P  be  the  pressure  exerted  by  the  fluid  upon  the  vane 
whose  velocity  is  u,  the  rate  at  which  work  will  be  done — or 
the  Mechanical  Power  * — or  simply  the  Power — will  be 

Pu,  (299) 

where  P  will  be  the  value  given  by  equation  (295)  for  this 
case,  the  entire  work  according  to  the  supposition,  being  done 
in  the  line  of  x.     Hence  (299)  and  (295)  give 

Pu  =  Xeii  =  Mic  {V  -  u)  (1  -  cos  6),         (300) 

If  the  deflection  be  through  a  right  angle,  make  6  =  90° ; 
and  if  two  right  angles,  make  d  —  18^. 

*  Author's  Elementary  Mechanics,  Article  90. 


324  REACTION  OF  FLUIDS.  [212.] 

Equation  (300)  is  a  maximum  in  reference  to  'w  as  a  varia- 
ble, when 

u  =  iv,  (301) 

which  reduces  (300)  to 

Pu  =  Xfl .  it;  =  i  Mv^  (1  -  cos  0),  (302) 

and  wlien  the  deflection  is  90°,  this  becomes 

Pic  =  AV .  iv  =  iMv^ ;  (303) 


and  for  180'^ 


Pu  =  X^.iv  =  iMv\  (304) 


Equation  (304)  gives  an  amount  of  work  equal  to  the  entire 
energy  of  the  stream  ;  if  a  fluid  stream  flows  into  a  vane,  like 
Fig.  158,  or  Fig.  162,  where  the  fluid  veins  are  completely  re- 
versed in  direction,  and  the  vane  is  urged  forward  with  half 
the  velocity  of  the  stream,  the  mechanical  power  imparted  to 
the  vane  will  equal  the  entire  energy  of  the  stream,  and  the 
effiowncy  is  said  to  be  j9^/y^c^.  In  this  case  the  fluid  leaves  the 
vane  with  no  actual  velocity. 

212.  We  will  now  deduce  these  results  from  the  principle  of 
the  Conservation  of  Energy.     There  being  no  loss  of  energy 

from  friction  or  otherwise,  the 
kinetic  energy  of  the  stream  be- 
fore entering  the  vane  will  equal 
the  energy  at  discharge  plus  the 
work  imparted  to  the  vane  ;  or 


Fig.  167. 


iMv"  =  Pu  +  iMV\  (305) 


where  T^is  the  velocity  of  discharge. 

Let  FG  be  the  stream  having  a  velocity  v,  GP  the  direction 
of  motion  of  the  vane  having  a  velocity  u  in  the  same  direc- 
tion as  the  stream  ;  then  will  v  —  u  be  the  velocity  of  the 
stream  relatively  to  the  vane  at  entrance,  and  since  there  are 


[218,  214.]  WORK  OF  IMPULSE,  325 

no  resistances,  it  will  quit  the  vane  tangentially  with  the  ve- 
locity BD  =  v  —  u,  and  at  the  same  time  it  will  move  for- 
ward with  the  velocity  BC  =  w ;  hence  the  actual  velocity 
will  be 

BE=  V=  V(v  -  uf  ^u^  ^%u(v-  u)  cos  6  ;    (306) 

which  in  equation  (305)  gives 

^Mv"  =  Pu  +  iM[{v  -  uy  +  u^  +  2u{v-  u)  cos  6]  ;     (307) 

from  which  we  find, 

I^u  =  Mu  (v  —  u)  {1  —  cos  ff), 

which  is  the  same  as  equation  (300).  Dividing  by  u  gives 
equation  (295).     From  (305)  we  have 

Pu  =  \Mv^  -  i  MV\  (308) 

or,  Th£,  energy  irrvparted  to  the  vane  equals  the  loss  of  energy 
of  tfie  fluid, 

213.  Efficiency. — Tlie  efficiency  of  a  stream  in  imparting 
work  to  a  vane,  is  the  ratio  of  the  energy  so  imparted  to  the 
axitual  energy  of  the  stream. 

Let  e  be  the  efficiency ;  then,  for  the  preceding  case,  we 
have,  equation  (308), 

'  =  im=  — im —  =  ^  - 1?-'    (^^^) 

214.  The  Resultant  Pressure  in  Fig.  167  will  be 


B  =  VXIVY\=  M{v  -u)2  sin  ^6 ;        (310) 
and 

tan  BGP  =  --'  =  cot  id  =  tan  (90°  -  iff) ; 
.-.  BGP  =  K180°  -0)  =  iFGB.  (311) 


826 


WORK  OF  IMPULSE. 


[215.] 


The  line  of  the  resultant  pressure  bisects  the  angle  hetween 
the  ajyproaching  stream  FG  and  the  plane  AB, 

215.  General  Case. — Let  the  vane  move  in  any  direction. 
Let  AB  be  the  vane,  CD  the  initial  direction  of  the  stream, 
having  a  velocity  v  =  DF,  u  =  DE~  the  velocity  and  direc- 
tion of  motion  of  the  vane ;  v^  =  BG 
=  the  velocity  of  discharge  relatively 
to  the  vane,  which  will  be  tangentially 
to  the  vane;    V  =  BII  =    the  actual 
velocity  of  discharge,  and  P  the  press- 
ure exerted  in  the  direction  of  motion 
DF.      There  being  no  frictional   re- 
H  sistance,  w^e  have,  as  before, 


Fig.  168. 


IMv^  =  Pu  +  iMV\         (312) 


To  find  F,  let  6  =  FDJ=  the  angle  between  the  stream 
prolonged  and  a  tangent  to  the  vane  at  the  point  of  impulse ; 
(p  =  FDF,  /3  =  the  angle  between  DF  iind  the  tangent  BG. 
The  line  ^i^  joining  ^and  F,  represents  the  magnitude  and 
direction  of  the  stream  relatively  to  the  vane  at  Z>,  and  since 
there  is  no  loss  of  velocity  relatively  to  the  vane  while  passing 
along  it,  we  have 


BG  =  EF  =Vi=  Vv^  +  u""  -  2vu  cos  <p,       (313) 

Drawing  GH  equal  and  parallel  to  DF  =  u,  the  line  BH 
=  V,  will  represent  the  actual  velocity  of  discharge,  and  we 
have 

Y^  =  v\  +  u^  -  2v^u  cos  BGH, 

=  v\-\-u^  +  2viU  cos  (/?  +  ^) ;         (314) 

which  after  substituting  Vi  from  equation  (313)  gives 


Y^  =  v^  —  2u  (v  cos  (p  —  u) 


+  2u  cos  (  /?  +  ^  )  \/v^  ■\-  u^  —  2vu  cos 


-  \ ;    (315) 


t^^^-^  AND  PRESSURE. 

hence,  equations  (312)  and  (315)  give 
Pu  =  Mu[v  cos  cp 


327 


u 


—  cos  (^  +  (p)  \/'i^  +  u^  —  2vu  cos  (p]  ; 

',  P  =.  M{v  cos  (p  —  XL 

—  cos  (/?  +  9?)  V't^  +  '?^^  —  2vw  cos  q)) 
=  iff  [v  cos  (p  —  u  —  Vi  cos  (/?  +  ^)] 


(316) 
.(317) 


If  k  be  the  loss  of  energy  compared  with  the  total  energy 
of  tlie  stream,  the  efficiency  in  imparting  work  to  the  vane 
will  be 


e=l 


Pu 


cos  cp 


u' 


\  ;  (318) 


-  COS  (/J  4-  9>)^1  +  -^  -  2-eos  9.  J 


■from  which  the  condition  for  maximum  efficiency,  u  being 
variable,  may  be  found,  but  the  result  will  be  too  complex  to 
be  of  practical  value.  These  equations  include  not  only  all 
the  results  of  the  preceding  cases,  but  also  several  others. 
They  are  independent  of  the  initial  slope  ^  of  the  vane,  rela- 
tively to  the  stream,  but  are  dependent  upon  the  relative  di- 
rection /?  with  which  it  quits  the  vane. 

216.  Discussion. — Case  T.  Let vco^cp  —u=—Vi cos (^  +  cp). 
Then   equations  (317),  (316),  (315),  (318)  become,  respect- 
ively, 

P  =  2M{vco8  tp  — u) 

Pu  =  2M{v  cos  (p  —  u)u 

F^  =  'y'  —  4  (v  cos  (p  —  u)u  r  •  (319) 

_  4  (vcos  q)  —  n)u 


328  WORK  OF  IMPULSE  [216.] 

In  this  case  v  cos  cp  is  the  projection  of  the  velocity  v  on  the 
direction  of  motion  of  the  vane-w;  and  in  all  cases  where  work 
is  imparted  to  the  vane,  the  latter  will  be  less  than  the  for- 
mer, and  hence  v  cos  cp  —  u  will  be  positive,  and  cos  {/3  +  (p) 
will  be  negative,  ov  fi+  cp  will  exceed  90°.  The  second 
member,  v^  cos  (/?  +  ^),  is  the  projection  of  v^  on  the  direc- 
tion of  u\  hence,  the  assumption  makes  the  resultant  ]3ressure 
coincide  w^ith  the  direction  of  motion  of  tlie  vane ;  also  the 
projection  equals 

EFqosDEF\ 
hence, 

cos  KEF=  —  cos{(p  +  p); 

,'.  KEF  =  180'^  -{cp  +  /3).  (320) 

The  internal  angles  ^at  Z>  and  iV"  of  the  triangle  EDN 
equal  the  external  angle  KEN^  or 

KEN  =(p  +  ft', 

.-.  DEN  =  180"  -  KEN  =  180°  -{cp  +  /3);       (321) 

hence  (320),  (321), 

DEN=KEF, 

as  it  should,  since,  as  shown  above,  Vi  may  be  laid  off  on  EF 
or  EN;  and  if  the  angles  be  all  measured  from  Z>^  prolor^ed, 
we  have,  equation  (321) :  The  angle  hetween  the  direction  of 
^notion  of  the  vane^  and  that  of  the  stream  relatively  to  the  vane 
at  the  pohit  of  impulse^  must  equal  the  sujyplenient  of  the  angle 
hetioeen  the  former  line  and  that  of  the  direction  of  the  stream 
relatively  to  the  vane  where  it  quits  it. 

The  line  GB  produced  will  not  generally  pass  through  E, 
but  in  any  case  the  relation  of  the  angles  will  be  that  given 
above,  since  a  line  may  be  drawn  through  ^parallel  to  BG. 

The  speed  of  maximum  efficiency  will  be  found  by  making 
the  second  member  of  the  last  of  equations  (319)  a  maximum 
in  reference  to  ^^  as  a  variable,  which  requires  that 

u  =  ivQo^cp]  (322) 


[216.  J  AND  PRESSURE. 

which  reduces  (319)  to 

P  =  Mv  cos  q) 

Fu  =  iMv^  cos'  (p 
Yz=  V  sill  (p 
e  =  cos*^  (p 


829 


(323) 


and  P,  Puy  and  e  will  be  a  maximum  for  (p=  0,  when  F  will 
be  zero. 

Case  II.  Let  the  vane  be  flat  and  oblique  to  the  stream. 

For  this  case  /?  becomes  6^  and  by  substituting  the  latter  for 
p  in  (315),  (316),  317),  (318),  tlie  required  results  will  be  found. 
The  equations  will  be  of  the  same  form  as  for  the  general  case. 

Case  III.  Let  the  vane  be  flxit  and  move  normally  to  the 
Wine, 

For  this  case  ^  =  /?,  and 

^  +  /?  =  90°  ;  ^       (324) 

and  equations  (315),  (316),  (317),  (318)  become 

P  =  M{v  cos  cp  —  u) 

Pu  =  M{v  cos  q)  —  u)u 
V'  =  ^-  2u  (v  cos  <p-u)  r  •  (^2^) 

2(v  cos  (p  —  u)u 

'=- — ^ — - 

The  speed  for  maximum  efficiency  will  be  when 

w  =  |v  COS  ^,  (326) 


which  reduce  (325)  to 


330 


EFFICIENCY 
P  =  iMv  COS  (p 

Pu  =  \Mv'^  cos^  (p 


[216.] 


(327) 


e  —  ^cos^  cp 

Case  IY.  Zd^^  ^A^  vq^t?^  he  flat  and  normal  to  tJie  stream^  the 
vane  onoving  at  an  angle  cp  with  the  stream. 

In  tliis  case  fi  =  90°,  and  we  have  from  (317),  (316),  (315), 


P  =  M  {v  cos(p—  u  -{■  sin  (p  ^v^  +  u^  —  2vu  cos  ^) 


Pu  —  M{v  cos  (p  —  u  -\-  sm  cp  ^/v^  +  u^—  2vuco&  (p)i 
Y^  =  v^  —  2(v  cos  (p  —  u)u 


—  2u  sin  cp  A^v^  -\-  1^  —  2vu  cos  cp 


.  (328) 


Case  Y.  Let  tJie  vane  he  flat  and  normal  to  the  stream^  and 
move  in  the  direction  of  the  stream. 

In  this  case  ^  ^  90°,   <7?  =  0,  which  in  (315)  to   (318),  or 
(325),  (326),  (327),  give 

P  =  M(y-u) 

Pu  =  M{v  —  u)u 
V''=v^-  2uv  +  2uf 


=  U^  +  (y  —  uf 

_  2(v  —  u)u 
^~  v^ 

and  for  maximum  efficiency, 

u=  \v 

P  =  iMv 

Pu^lMif' 

v = w 


(329) 


(330) 


[216.] 


OF  VANES. 


331 


Case  YI.  Lei  the  vaiie  he  a  heTnisphere  moving  in  the  direc- 
tion of  the  stream. 

Then  ft  =  180%        <p=0, 

and  is  a  sub-case  of  Case  I.,  hence  equations  (319)  to  (323)  be- 
come 

F  =  2M{v  -u) 


Pu  =  2M{v  —  u)u 
F^  =  (v  -  2uf 

4:(V  —  U)U 

and  for  maximum  efficiency 

u  =  iv 

P  =  Mv 

Pu  =  iM^  ^ 
V=0 
e  =  l 

Case  YII.  Zet  t/ie  vane  rnove 
normally  to  the  stream,  and  ft 
=  90°. 

Then  cp  =  90°, 

ft+  cp=  180°, 

and  we  have 

P  =  M{-  u  4-  V^  +  u") 


(331) 


(332) 


Pu  =  M{-  u  +  Vi^  +  u^)u 


e  = 


2(  —  u  +  V'l^  +  v^)u 
v" 


(333) 


This  is,  substantially,   a  sub-case  of   Case   lY.,  when    g? 
=  90°. 


332  HYDRAULIC  [217,  218.] 

HYDRAULIC   MOTORS. 

217.  A  hydraulic  motor  is  any  device  or  machine  by  means 
of  which  the  energy  of  water  may  be  utilized.  These  ma- 
chines are  of  various  classes,  among  which  we  notice  the  water 
pressure  engine,  in  which  the  water  passes  into  and  out  of  the 
machine  through  ports,  similarly  to  steam  in  the  steam  engine ; 
water  wheels,  in  which  the  water  flows  against  floats  at  the 
outer  circumference  of  the  wheel ;  turhines,  in  which  the 
water  passes  through  the  wheel  either  radially  or  parallel  to 
the  axis  of  the  wheel ;  and  reaction  wheels  in  which  the  wlieel 
is  actuated  by  tlie  reaction  of  the  water  passing  through  it. 

In  making  an  application  of  the  preceding  principles,  it  will 
be  necessary  only  to  describe  the  general  features  of  the  con- 
struction of  each  machine,  leaving  the  details  to  those  treatises 
which  make  a  specialty  of  this  subject.  That  part  of  the 
stream  which  impinges  against  the  vane  of  a  motor  will  be 
called  ?ijet  to  distinguish  it  from  the  stream  which  supplies  the 
reservoir. 

218.  A  SINGLE  VANE  moviug  with  a  velocity  u  normally 
before  a  jet  having  a  velocity  v,  will  be  impinged  upon  with 
the  relative  velocity  v  —  u,  and  hence  the  quantity  Impinging 
per  second  wull  be 

mlc  {v  —  n), 

and  the  relative  momentum  will  be 

P  =  'L],,{^-uf.^  (334) 

y 

and  the  work  done  on  the  vane  will  be 

Pu  =  —Mv-ufu',  (335) 

which  is  a  maximum  for 

u=  \v,  (336) 

It  will  be  observed  that  this  analysis  gives  the  same  result 


[219-221.] 


MOTORS. 


388 


as  substituting  — k  {v  —  u)  for  Mm  equation  (300),  and  making 

if 

0  =  90°.  The  analysis  just  given  assumes  that  only  a  portion 
of  the  water  passing  a  given  fixed  section  of  the  jet  is  utilized 
in  producing  work,  as  will  be  the  case  when  only  a  single 
vane  receives  the  impulse.  A  single  vane  will  not  be  used  in 
practice,  but  instead  thereof  a  succession  of  vanes  at  short  in- 
tervals ;  in  which  case  it  is  assumed  that  the  entire  mass  of  the 
water  in  the  jet  is  directly  involved  in  producing  work. 


VERTICAL    WHEELS. 

219.  Vertical  wheels  are  such  as  revolve  in  a  vertical 
plane,  and  hence  their  axes  of  rotation  are  horizontal.  The 
principal  classes  are  Undershot  Wheels,  Breast  Wheels^  and 
Overshot  Wheels. 


lias  vanes. 


floats,  or  buckets 


220.  The  undershot  wheel 
at  its  circumference  for  receiv- 
ing the  water.  The  water  flows 
under  the  wheel  nearly  horizon- 
tally, and  after  doing  its  work 
upon  the  vanes  escapes  through 
the  tail-race.  There  arc  two 
kinds — one  witli  flat  vanes,  the 
other  with  curved  vanes. 


Fio.  170. 

221.  In  undershot  wheels  with  flat  vanes  it  is  assumed 
that  the  water  impinges  npon  the  vanes  normally  and  leaves 
them  tangentially ;  which  suppositions  being  very  nearly  re- 
alized, are  sufficiently  accurate  in  practice,  and  make  equations 
(329)  and  (330)  directly  applicable. 


Let 


Q  =  the  number  of  cubic  feet  per  second  of  the  water 
discharged  by  the  jet  against  the  vanes  ; 


334  WATER  WHEELS.  [221.] 

Then  will 

62iQ  =  the  pounds  of  water  discharged  per  second  ; 

and 

M  =  -5||^  =  2Q  approximately,  (337) 

and  the  mechanical  power  imparted  to  the  wheel  will  be,  2d 
of  Equations  (329), 

Pu=.^Q{v-u)u;  (338) 

which  may  be  written  : 

The  three  terms  within  the  parenthesis  are  expressions  of 
heads  due  to  velocities,  and  their  algebraic  sum  is  called  the 
effective  head  /  hence 

The  effective  head  is  tlwhead  due  to  the  velocity  of  entrance  of 
the  jet  LESS  the  head  due  to  the  velocity  in  the  wheel  at  exit,  and 
«?so. LESS  the  head  lost  at  the  entrance  of  the  jet  into  the  vane, 
and  the  two  latter  must  he  a  minimum  that  the  effective  head 
shall  he  a  maximum. 

This  principle  may  be  generalized  and  used  in  all  cases; 
thus :     Let 
II  =  the  total  available  head  of  the  fall, 
h  =  the  head  due  to  changes  of  velocity  at  the  entrance  of 

the  float ; 
h'  =  the  head  due  to  the  velocity  with  which  the  water  quits 

the  wheel  ; 
h"  =  the  head  due  to  frictional  resistances ; 
h'"  —  the  head  due  to  whirls  or  other  causes. 

Then 

Pu  =  mQiH-  h-h!  --K'  -  A"'] ;      (338  J) 


[321.]  WATER  WHEELS.  335 

and 


e  =  ^,  (3880) 

the  horse  power  developed  will  be 

60  X  m     Q^^_  ^)^  (339) 

32i  X  33,00U  ^  ^  ^    '  ^       ^ 

which  will  be  theoretically  a  maximum  when  the  circumfer- 
ence of  the  wheel  has  one-half  the  velocity  of  the  jet ;  in 
which  case  (339)  becomes 

^^^Z'l\x  2,200^''''         / 


^       nearly, 


1,132 


=  I  (2A,       ^  (340) 

where  h  is  the  head  due  to  the  velocity  v.  If  S  be  the  area  in 
feet  of  the  section  of  greatest  contraction,  we  have 

Q=.vS=S^/2gh'y 

,\JIP  =  0.4c55SShl  •  (341) 

The  theoretical  maximum  efficiency  in  this  case  will  be  (5th 
of  equations  (330)) 

but  this  considerably  exceeds  the  value  realized  in  practice. 
The  several  losses  due  to  the  contraction  of  the  vein,  the  clear- 
ances about  the  wheel  through  which  the  water  escapes  with- 
out acting  upon  the  wheel,  the  lack  of  normal  impulse,  the 
imperfect  action  of  a  thick  vein,  and  other  causes,  combine  to 
reduce  the  theoretical  efficiency.     Experiments  show  that  a 


336 


WATER  WHEELS. 


[223.] 


well-made  wheel  will  realize  about  60  per  cent,  of  the  theoret- 
ical efficiency,  giving  for  fair  practice,  equation  (339), 

For  a  practical  maximum  efficiency  we  have 

ei  =  from  0.30  to  0.3S.  (343) 

The  power  oi  the  wheel  is  independent  of  its  size  ;  and  hence 
may  be  so  proportioned  as  to  make  a  desired  number  of  revo- 
lutions per  minute. 

222.  The  Poncelet  Wheel. — M.  Poncelet,  a  celebrated 

French  scientist,  improved  the 
imdershot  wheel  by  making 
the  vanes  so  curved  that  the 
water  upon  leaving  them 
would  flow  backward  rela- 
tively to  the  vane.  For  this 
case,  equations  (331)  and  (332) 
are  applicable,  and  we  have 
theoretically,' 


HP 


60  X  62i 
321x33,000 


Q{v  —  u)u ; 


(344) 


and  for  a  theoretical  maximum 


IIP 


^6  """^^y- 


(345) 


If  the  velocity  of  the  circumference  of  the  wheel  be  |-  that 
of  the  jet,  and  the  buckets  be  so  curved  as  to  completely  re- 
verse the  direction  of  motion  of  the  water,  the  wheel  should 
be  perfect,  or 

e=l,  (346) 


but  these  conditions  not  being  realized,  combined  with  the 


[223,  224] 


WATER   WHEELS. 


337 


losses  noticed  in  the  preceding  Article,  make  the  prcLctlcal  ef- 
ficiency in  good  wheels  about 


0.60. 


(347) 


223*  A  BREAST  WHEEL  is  one  in  which  the  water  is  admitted 
at  some  point  opposite  the  face  of  the  wheel,  and  the  water 
retained  in  the  wheel  to  the  lowest  point  by  means  of  a  curved 
trough,  or  passage  way.  The  jet  will  enter  the  wheel  with  a 
velocity,  hut  if  the  buckets  be  properly  curved  and  the  velocity 
of  the  wheel  be  properly  regulated,  there  will  be,  theoretically, 
no  loss  from  tliis  cause. 
After  the  water  has  entered 
the  wheel  it  will  act  by  its 
weight  through  the  remain- 
ing height.  Let  h  be  the 
liead  due  to  the  velocity, 
and  ^1  =  CD  =  the  height 
through  which  the  water 
acts  by  its  weight ;  then 
will  the  maximum  theoretical  power  be 


Fig.  178. 


62je(A  +  Ai). 


(348) 


If  77=  EF^m  the  entire  fall  of  the  water,  the  theoretical 
power  will  be 

62  J  e^. 


and  hence  the  iheoreticdl  efficiency  will  be 

h  +  7/1 


e  = 


// 


(349) 


Experiments  with  the  best  wheels  of  this  class  have  given  an 
actual  efficiency  of 

ei  =  .75  to  .80.  (350) 

224.  Overshot  wheels  receive  the  water  at  or  near  their 
highest  part,  and  retain  it  in  buckets  during  its  descent.     If 

ad 


338 


EXAMPLES. 


[224.] 


retained  only  by  buckets,  the  water  spills  out  before  reaching 
the  bottom  of  the  wheel,  and  thus  produces  loss  of  efficiency ; 

and  if  the  water  be 
retained  by  a  curved 
trough,  this  class  of 
wheels  will  not  dif- 
fer essentially  from 
the  breast  wheel  in 
theory. 

If  the  water  acts 

by    its    weight 

through  the  height 

^     ,,,  -  CD  =  ]h  then   will 

Fig.  173 

the  mechanical  power  be 


Pu  =AM{v  -  u)u  +  62iQhi, 


(351) 


where  ^  is  a  coefficient  whose  value  in  the  case  of  flat  vanes 
will  be  unity,  as  shown  by  the  2d  of  Equations  (329),  and  for  a 
complete  reversal  of  direction  of  the  jet  will  be  2,  as  shown 
by  the  2d  of  equations  (331) ;  hence  the  real  value  of  A  will 
be  more  than  1  and  less  than  2. 

If  the  velocity  of  the  circumference,  n^  of  the  wheel  be  half 
that  of  the  water  when  it  enters  the  wheel,  and  there  be  no 


A{v  —  n)  u 


26- 


loss  from  the  impulse,  then 
and  we  have 


will  equal  AB  =  h 


(352) 


the  same  as  (342),  and  the  efficiency  of  the  wheel  would  be 
perfect.  But  there  will  necessarily  be  some  loss  from  impulse, 
and  an  additional  amount  from  clearance  in  the  curved  trough, 
and  still  more  from  the  imperfection  of  action  due  to  the 
thickness  of  the  jet,  all  of  which  combine  to  make  the  practi- 
cal efficiency  of  wheels  well  constructed  and  properly  oper- 
ated, of 


ei  =  0.75  to  0.80. 


(353) 


[224]  EXAMPLES.  889 

•  EXAMI'LES. 

1.  Water  approaches  an  overshot  wheel  with  a  velocity  of  12 
feet  per  second,  the  buckets  moving  with  a  velocity  of  6  feet 
per  second  where  the  fall  is  20  feet,  but  on  account  of  empty- 
ing the  buckets  4  feet  are  wasted  ;  find  the  efficiency,  and  if 
the  supply  be  600  cubic  feet  per  minute,  find  the  horse  power. 

The  relative  velocity  of  6  feet  per  second  in  the  race  is  equiv- 

6^ 
alent  to  a  head  li  =  — .     The  height  through  which  the  water 

acts  by  its  weight  will  be  20  —  4  =  16  feet. 

Then,  for  useful  work  per  second,  we  have,  making  A  =  1 
in  equation  (351), 

^,mQ{12  -6)  X  6  +  wQ  xl6, 
or 

36  —  ^+  IQwQ. 

The  energy  at  command  will  be 

imQ (12f  -{-2010 Q, 


or 


and  we  have  for  the  efficiency, 


36  4-  16  X  32^  _   ^ 

72+  20  X  32J 

The  horse  power  w^ill  be 

(36  4-  16  X  32^ )  X  62^  x  600  _.^^ 
33,000  X  32i 

2.  The  section  of  a  streai^  of  water  falling  vertically  over  a 
dam  12  feet  high  is  one  square  foot  at  the  foot  of  the  fall ;  re- 
quired the  horse-power.  An8,  37.87  + . 


340  EXAMPLES.  [224.1 

3.  If  a  breast  wheel  whose  diameter  equals  the  height  of  fall, 
receives  a  stream  of  water,  section  /S',  directly  opposite  its  centre, 
if  there  be  a  loss  of  5  per  cent,  of  head  on  account  of  the  clear- 
ance at  the  bottom  of  the  wheel,  and  of  10  per  cent,  of  velocity 
at  the  contracted  section  ;  required  the  horse  power  and  the 
maximum  efficiency. 

If  the  vanes  be  flat,  and  the  impulse  normal,  the  velocity  of 
the  circumference  beini^  ?/,  equations  (329)  will  be  applicable, 
but  in  properly  constructed  vanes,  Case  Y.,  equations  (331) 
will  be  applicable.  Assuming  the  latter,  let  Ji  be  the  radius 
of  the  wheel,  then 


V  =  O.yO  ^/^</A^ 
and,  for  maximum  efficiency  u  =  ^v,  or 
n  =  0.45  ^2^ ; 
and  the  power  of  the  impulse  will  be  (3d  of  (332)), 

2(/ 
the  work  done  by  the  weight  will  be 

621  X  ^  X  0.95^; 
hence  the  total  work  will  be 

110  QB.  (a) 

The  work  done  by  62  J  ^  falling  a  height  27?  will  be 
2  X  (j2iQJ2; 
hence  the  maximum  efficiency  will  be 


[324.^  EXAMPLES,  341 

To  find  the  horse  power,  substitute 

in  equation  (a),  and  find 

110  X  7.2  V^ .  x^  X  i2  X  60 


{0) 


HP 


33,000 


=  1.44^^^'. 


4.  Required  the  horse  power  and  efficiency  of  an  under- 
shot wheel  when  the  velocity  of  its  circumference  is  \  that  of 
the  jet ;  the  quantity  of  water  discharged  being  600  cubic  feet 
per  minute. 

5.  Required  the  horse  power  and  efficiency  of  an  undershot 
wheel  when  the  velocity  of  its  vanes  is  J  that  of  tlie  jet,  when 
Q  =  600  cubic  feet  per  minute. 

G.  If  the  discharge  of  water  against  a  Poncelet  wheel  be  100 
cubic  feet  per  second,  and  the  velocity  of  its  perimeter  be  { 
that  of  the  jet,  required  the  horse-power  and  efficiency. 

7.  Determine  the  horse  power  and  efficiency  in  the  preced- 
ing example  if  the  velocity  of  its  perimeter  be  f  that  of  the 
jet. 

For  the  horse  power  we  have,  equation  (344), 

-^^  =^32ix  33,000  '^^^(^-^^)^^> 
=  -^t^  nearl3^ 
For  the  efficiency  we  have 

_  2M{v  -  iv)iv  _ 
^~  iM?        ""^• 

8.  If  the  velocity  of  the  perimeter  of  a  Poncelet  wheel  be  6 


342  EFFECTIVE  HEAD.  [225.] 

feet  per  second,  and  the  discharge  of  the  jet  be  200  cubic  feet 
per  minute ;  determine  the  section  of  the  jet  for  maximum 
efficiency. 

9.  In  the  preceding  example,  determine  the  section  of  the 
jet  when  the  efficiency  is  one-fourth  the  theoretical. 

10.  In  the  Poncelet  wheel,  the  velocity,  u,  of  the  circumfer- 
ence being  fixed,  and  the  required  horse  power,  ^-P,  being 
given  ;  find  the  head  due  to  the  velocity  when  the  efficiency  is 
a  maximum,  and  the  section  of  the  orifice,  the  coefficient  of 
contraction  of  the  vein  being  0.62. 

11.  At  what  velocities  of  the  wheel  will  the  efficiency  be 
zero? 

12.  If  a  wheel  is  being  run  at  a  velocity  to  produce  a  maxi- 
mum efficiency  with  a  jet  of  25  square  inches,  how  much  must 
the  section  of  the  jet  be  increased  to  produce  the  same  work 
when  the  wheel  revolves  with  |  the  former  velocity  ? 

225.  It  will  be  seen  from  the  preceding  discussion  that  if 
the  water  can  be  made  to  enter  the  wheel  or  act  upon  its  vanes, 
without  shock  and  leave  it  without  velocity,  the  highest  ef- 
ficiency will  be  produced  ;  and  if  in  addition  to  these  there  be 
no  losses  from  friction  or  clearance,  or  other  imperfect  condi- 
tions, the  efficiency  will  be  unity,  and  the  wheel  is  said  to  be 
perfect. 

Some  writers  distinguish  between  impulse  and  reaction  / 
the  former  being  applied  to  the  direct  action  of  the  water  in 
producing  the  impulse;  and  the  latter  to  the  effect  due  to 
the  change  of  motion  after  the  water  has  entered  the  vane,  and 
hence  maybe  affected  by  friction  and  the  direction  with  which 
it  leaves  the  vane.  According  to  the  view  presented  above, 
the  action  of  liquids,  whether  of  impulse  or  pressure,  is  of  the 
nature  of  a  reaction. 

T/ie  effective  head  is  that  portion  of  the  actual  head  which 
would  produce  the  work  done  by  acting  upon  the  wheel  with- 
out loss  of  any  kind.  Tims  the  coefficient  of  M  in  equations 
(300),  (316),  (323),  (325),  etc.,  divided  by  g,  and    {v  -  n)  u 


[226.]  REACTION   MOTORS.  343 

-^  g  in  equations  (338)  and  (344)  are  effective  heads  for  the 
respective  cases  here  cited. 

REACTION   MOTORS, 

226.  Jet  Propeller. — If  water  issues  horizontally  from 
the  side  of  a  vessel,  it  will,  by  virtue  of  its  reaction,  move  tlie 
vessel  in  the  direction  opposite  to  that  of 
the  jet,  when  the  resulting  pressure  is  suf- 
ficient to  overcome  the  resistances.  As- 
suminor  that  there  are  no  frictional  resis- 


tiances,  the  case  will  be  similar  to  Case  ^^^^mm^M^ 
TIL  of  Article  216.  ^'^■'^'^^ 

A  jet  issuing  from  an  orifice  having  a  component  motion 
either  with  or  opposite  to  that  of  the  jet,  will  have  a  velocity 
relatively  to  the  orifice  equal  to  the  component  velocity  of  the 
orifice,  abstraction  being  made  of  the  velocity  due  to  the  head 
of  the  liquid  in  the  vessel ;  just  as  a  body  on  a  smooth  plane 
in  motion  will  have  a  velocity  in  reference  to  the  plane  inde- 
pendently of  the  velocity  of  the  body  in  reference  to  the  earth. 

Let  h  =  AB  =  the  static  head  above  the  orifice  B, 
V  =  the  velocity  of  discharge  due  to  the  head  A, 
u  =  the  velocity  of  the  vessel, 
hi=  the  head  due  to  the  velocity  w, 
V=  the  velocity  of  discharge  relatively  to  the  orifice; 
then, 

v'  =  2gh,  u"  =  2y/iu  (354) 

and 

V  =  2^  (A  +  Ai)  =  ^^  +  u\  (355) 

The  actual  velocity  of  the  jet  will  be 

V-  u', 

which  will  be  less  than  v  for  any  finite  value  of  t/,  and  the 
pressure  due  to  the  reaction  will  be,  equation  (297), 


P  =  .¥(  F-  'w)  =  Jf  (V^  +  u?  -  u),        (366) 


844  EEACTION  MOTORS.  [227.] 

which  may  be  compared   with   the  first  of  equations  (333). 
The  mechanical  power  will  be 


Pu  =  M{^u^  +  ?/  -  u)  u,  (357) 

which  has  no  theoretical  maximum.     Developing,  we  have 

p..  =  i;T/(.^-£..  +  |,-|^,+  ete.),     (358) 

which  is  iMi/  for  u  =  oo,  for  which  case  the  motor  would  be 
perfect.  But  it  is  impossible  to  realize  tliis  condition,  and  it 
is  found  in  practice  tliat  the  efficiency  is  a  maximum  for  u 
equal  nearly  to  v.     Making  u  =  v,  in  (357),  we  have 

Fu  =  Jf  (a/2  -  1) v''  =0AU2i¥i/.  (359) 

The  potential  energy  of   the  water  having  the  head  AB 
will  be 

e2lQ'AB  =  i3Iv'',  (360) 

hence  the  theoretical  efiiciency  of  the  motor  will  be 


_M{Vw'  +  v^-  it)u 

=  2-=^L =  T^;  (361) 

in  which,  ii  v  =  u,  we  have 

€  =  0.82+.  (362) 

A  vessel  floating  on  water  might  be  propelled  according  to 
this  principle.  A  pump,  drawing  water  from  the  bow  of  a 
vessel  and  forcing  it  out  at  the  stern,  would  propel  the  vessel ; 
and  for  velocities  u>  v  would  be  theoretically  quite  efficient. 

227.  Barker's  (or  Seguin's)  Mill  consists  of  a  vertical 
hollow  shaft   communicating   with   hollow   transverse   arms. 


[227.]  REACTION  MOTORS.  345 

The  water  is  admitted  into  the  npper  end  of  the  hollow  shaft, 
and  passing  downward,  flows  horizontally  into  the  hollow 
arms,  escaping  horizontally  through  orifices  near  their  extrem- 
ities, one  orifice  being  on  one  side  of  one  arm,  and  the  other 
on  the  opposite  side  of  the  other  arm.  The  deflection  of  the 
water  from  the  vertical  to  the  horizon- 
tal direction  will  cause  both  a  hori- 
zontal and  vertical  pressure,  as  shown 
in  Article  21 ;  but  the  horizontal 
pressures  will  neutralize  each  other, 
while  the  vertical  pressure  will  be 
resisted  by  the  support,  and  the  water 
will  flow  into  the  arms  with  a  velocity 
unaffected  by  these  pressures,  and,  if  ^ 
the  arms  are  stationarj-,  the  water  will  ~-^^^^^^^^^^^f 
be  discharged    at   the  orifices  with  a  yiq.  n5. 

velocity  due  to  the  head  in  the  vertical,  shaft.  But  when  the 
arms  are  rotating,  the  water  in  them  will  be  forced  radially 
outward  on  account  of  the  centrifugal  force  developed,  and 
thus  the  head  of  discharge  relatively  to  the  orifice  will  be 
increased.  Or,  instead  of  considering  it  in  the  liglit  of  cen- 
trifugal force,  we  may  consider  that  the  water  has  a  velocity 
common  with  the  arm,  which  alone  would  be  equivalent  to  a 
head  producing  a  velocity.  As  the  water  approaches  the 
ends  of  the  arms  it  will  be  deflected  through  an  angle  of  90°, 
thus  producing  a  pressure  radially  outward,  and  also  trans- 
versely to  the  arms,  the  former  of  which  will  be  resisted  by 
the  solid  parts  of  the  machine,  and  the  latter  will  produce  the 
rotary  motion.  The  two  reactions — one  at  each  end  of  the 
outer  extremities  of  the  arms — ^produce  a  couple,  the  moment 
of  which  will  equal  the  moment  of  the  resistances  overcome 
by  the  machine. 

Let  V  =  the  velocity  due  to  the  head  h  =  OBy 

a  =  the  arm  to  the  orifice, 

CO  =  the  angular  velocity  of  the  arms, 

ti  =  the  actual  velocity  of  the  orifices  G  and  -?'=  aco, 

V  =  the  velocity  of  discharge  relatively  to  the  orifica 

hi  =  the  head  due  to  the  velocity  u. 


346  REACTION  MOTORS.  '  [227.] 

Then,  as  shown  in  the  preceding  Article, 

and  the  actual  velocity  of  discharge  will  be 

V-u=V'^^~+^^-aGD;  (364) 

and  the  pressure  due  to  the  reaction  will  be 


F  =  M  {Va^od'  +  v"  -  aoS)  ;  (365) 

and  the  mechanical  power  resulting,  will  be 


Fu  =  3f{Va''o^  +  v^-  aoo)  aoo ;  (366) 

and  the  theoretical  efficiency  will  be 


M^^d^oo^  +  zr^  —  aGD)aGo 


F+  ace?' 


(367) 


which  has  no  maximum  in  reference  to  g?  as  a  variable,  but 
approaches  unity  as  a  limit  as  qd  is  increased  indefinitely.  In 
practice  it  is  found  that  the  efficiency  is  a  maximum  when 
aos  is  about  equal  to  v  ;  and  making  aoo  —  v,  and  neglecting 
all  losses,  except  that  due  to  a  loss  of  actual  velocity  of  dis- 
charge, we  find  for  the  efficiency, 

^  =  0.82  4-.  (368) 

If  there  were  several  orifices  at  distances  a,  a",  a'",  -etc., 
from  the  axis  of  the  shaft,  producing  reactions  P\  P",  P"\ 
etc.,  giving  expressions  similar  to  those  in  equation  (365),  and 
Ji  the  resistance  overcome  w^ith  a  uniform  velocity  w  at  sl  dis- 
tance J  from  the  axis  of  the  same  shaft,  then  would  the  mo- 


[227.J  REACTION  MOTORS.  347 

ments  of  the  reactions  equal  the  moment  of  the  resistance 
overcome,  and  hence 

nh  =  Fa'  +  P"a"  +,  etc., 


=  MWa'^oj'  -\-v^-  a' go)  a' 


+  M"  {'Va"'^a^  +  1^-  a" GO)  a"  +  ,  etc. ;     (369) 
and  the  mechanical  power  will  be 


Moo  =  31'  (Va'^o^  -\-i^-  a' go)  a' go  +,  etc.         (370) 

An  analysis  of  the  second  member  of  the  last  equation 
gives  the  following  : — the  quantity  \^a"^Gj^  +  v^  —  a' go  is  the 
actual  velocity  of  discharge  as  shown  by  equation  (364) ; 
M'  (Va'^ca^  +  v^  —  a' go)  is  the  momentum  of  the  mass  M',  as 
shown  by  the  definition  in  Article  27;  and  M'Wa'^GcP-  +  i^ 
—  a  go)  a'  is  the  moment  of  the  momentum,  as  shown  in  Ar- 
ticle 166,  or  in  Article  168  of  the  Elementary  Mechanics. 
Hence  the  entire  expression,  M' {^a^G^  -v  i?  —  a  go)  a' go,  is 
the  moment  of  the  Tnomentuvi  of  the  mass  M'  multij)lied  by 
the  angular  velocity,  and  similarly  for  the  other  termr^.  Separat- 
ing the  expression  into  other  terms,  we  have  M{Va^Gj^  +  i^)a', 
which  is  the  moment  of  the  momentum  of  the  mass  M'  having 
the  velocity  Va'^co^  +  v^  m  reference  to  the  orifice ;  and 
M'a'Go-a,  which  is  the  moment  of  the  momentum  of  the 
mass  M'  having  an  actual  velocity  a'Go,  equal  to  the  actual 
velocity  of  the  orifice,  which,  being  in  a  direction  opposite  to 
that  of  the  discharge,  will  be  negative.  Each  of  these  mul- 
tiplied by  the  angular  velocity,  go,  will  give  the  corresponding 
mechanical  power  developed  ;  and  similarly  for  the  other 
terms  containing  M",  M'",  etc. 

In  tills  case  the  water  enters  the  arm  without  angular 
velocity,  still  the  principle  shown  in  Article  166  is  general, 
for  the  effect  produced  is  the  result  of  the  mutual  action  and 
reaction  between  the  water  and  arm.    Hence, 


848  REACTION   MOTORS.  [228.] 

When  water  hy  its  reaction  jproduces  a  rotary  motion^  the 
moment  of  momentum  into  the  angular  velocity  equals  the 
work  done. 

This  machine  has  been  the  subject  of  many  improvements. 
To  remove  as  much  as  possible  the  resistance  of  the  water  in 
passing  through  the  arms,  sudden  turns,  eddying,  etc.,  the 
arms  have  been  made  large  where  they  join  the  body  of 
the  sliaft,  and  gradually  contracted  and  curved  as  they  pass 
outward  toward  the  openings.  Instead  of  arms,  there  may 
^^_^  be  a  disc  having  openings  which  permit  the 

water  to  escape  tangentially,  which  form  is 
known  as  Wliitelaw's  Turbine.  In  the  form 
shown  in  Fig.  175,  the  pivot  at  the  lower  end 
of  the  shaft  supports  the  mill,  and  all  the 
water  running  through  it,  thus  producing 
Fig.  176.  mucli  friction.     This  objection  has  been  in 

part  removed  by  introducing  the  water  into  the  mill  from  the 
under  side,  thus  producing  an  upward  pressure,  and  counter- 
balancing, in  part,  the  weight  of  the  machine.  But  notwith- 
standing all  these  improvements,  the  mill  has  been  superseded 
by  other  more  efficient  motors. 

228.  To  determine  the  jpressure  due  to  the  centrifugal 
force, — Let  A  be  the  uniform  cross-section  of  the  ai-ms,  cd 
the  angular  velocity,  j?  the  pressure  per  unit,  m  the  mass  of  a 
unit  of  volume,  p  any  distance  from  the  axis  of  rotation,  the 
initial  and  terminal  values  of  which  are  r^  and  rg  respectively, 
then  will  the  mass  of  a  transverse  section  be 

7nAdpf 

and  the  centrifugal  force  will  be 

mAdp'  CO*/), 
which  will  equal 

Adj} ; 


Ad^  =  mAar^     pdp; 


.]  REACTION  MOTORS.  349 

or  ^  -  ^0  =  I mc^  {ii  -  r\),  (371) 

which  is  the  pressure  required. 

If  w  be  the  weight  of  a  unit  of  volume,  we  have 

^^  (J  -  f )  =  ^^  ('■  -  ^o)  =  -^  ('i  -  '  i)'        (372) 

where  h  is  the  head  due  to  the  velocity  7'^go,  and  //q  that  due 
to  TiGj.  The  velocity  due  to  the  difference  of  these  heads 
will  be 

u  =  V2f(h  -  /»o)  =  GoVA^i  ;  (373) 

in  which  if  Vi  =  0,  and  r^  =  a,  we  have 


ti  =  aoDs 


(374) 


which  is  the  same  as  u  in  equation  (363). 

But  both  the  velocity  of  the  orifice  and  centrifugal  force 
are  not  involved  in  determining  the  velocity  of  discharge. 

Centrifugal  force — or  energy — may  he  discarded  in  determining  the 
efficiency  of  a  motor  in  all  cases  where  the  head  due  to  the  centrifugal 
force  has  been  induced  by  the  rotation  of  the  wheel.  For  this  head 
will  have  heoip  secured  at  the  eocpense  of  the  energy  of  the  outflowing 
jet^  and^  in  turn,  this  head  vnU  impart  to  the  outflowing  jet  exactly 
the  same  amount  of  energy  as  that  which  it  received.  Or,  briefly, 
in  a  reaction  wheel,  or  turbine^  centrifugal  force  is  self  neutralized, 

1.  Find  the  horse  power,  number  of  revolutions  per  min- 
ute, gallons  of  water  discharged,  and  efficiency  of  a  Barker's 
mill,  neglecting  friction.  Area  of  nozzles  1  sq.  ft.,  radius  3 
feet,  total  head  30  feet,  speed  of  orifices  Jths  that  due  to  the 
head. 


u  =  J V2</  X  30  =  32.88  feet  per  second ; 
"inn  X  3  =  32.88  x  60, 
.-.  w  =  104.6; 


850  TURBINES.  [229.] 


V  =  Vi^2  ^  2(//i,  or  F=  54.8 ; 

^.        54.8  X  1,728  X  60      „ .  ^^, 
GU  = ~^ =  24,594. 

Efficiency  = pr  =  0.Y5. 

Horse  power  =  140.7. 

2.  A  rotating  wheel  receives  2  tons  of  water  per  minute 
at  a  radius  of  2  feet,  and  delivers  it  at  a  radius  of  3|  feet ;  on 
entrance,  the  water  is  rotating  with  a  velocity  of  14  feet  per 
second,  and  on  delivery  it  is  rotating  in  the  opposite  direction 
with  a  velocity  of  4  feet  per  second.  Find  the  couple  exerted 
on  the  wheel,  and  if  the  wheel  makes  200  revolutions  per 
minute,  lind  the  HP  developed. 

2  X  2240 
Moment  of  momentuvi  on  entry  =         ""LtT  x  14  x  2. 

bO    X   On 

2  X  2240 
Moment  of  momentum  on  delwery  =  -77^—^ — ^j^  x  4  x  3^. 

oU    X    oJi 

Sum  =  Coujple  =  98. 

„p      98  X  27r  X  200       ^  _ 
^-^  =  "    60  +  550       =^-^- 


TURBINES. 

229.  The  general  definition  of  a  turbine  is  a  water  motor 
rotating  about  a  vertical  axis ;  but  it  is  usually  restricted  to 
those  horizontal  wheels  in  which  the  water  is  conducted  to  the 
vanes  by  curved  guide-plates.  They  may  be  divided  into  four 
classes : 

1.  Outward  flow  turhines,  in  which  the  water  flows  hori- 
zontally outward  from  the  central  shaft,  and  is  discharged  at 
the  outer  circumference  of  the  wheel. 

2.  Inward  flow  turhines,  in  which  the  water  is  admitted  at 


[230.] 


TURBINES. 


351 


the  outer  circumference  and  flows  horizontally  inward  toward 
the  central  shaft. 

3.  ParaZld  flow  turbines^  in  which  the  water  flows  down- 
ward (or  upward)  through  the  wheel. 

4.  Mixed  flmo  turbines^  in  which  the  flow  may  be  com- 
pounded of  the  Ist  or  2d  with  the  3d  above. 

230.  Fourneyron's  Turbine,  invented  about  the  year  1827, 
is  a  good  type  of  an  outward  flow  turbine.     The  central  shaft 


Fio.  177. 


(7  of  the  wheel,  to  which  the  driving  mechanism  is  attaclied, 
passes  down  tlirougli  the  supply  chamber  GII^  from  which  it 
is  separated  by  the  tube  LL, 


352  TURBINES.  [280.] 

The  water  passes  downward  between  the  guide  plates  oo^ 
and  outward  between  the  same,  through  the  gates  MM^  thence 
against  and  along  the  vanes  AB,  finally  escaping  at  the  outer 
circumference  of  the  wheel.  The  vanes  curve  backward  so  as 
to  cause  the  water  to  be  discharged  in  a  direction  opposite  to 
that  of  the  wheel,  relatively  to  the  vanes.  The  vanes  are 
attached  to  the  shaft  C  by  means  of  the  rigid  arms jj^. 

To  find  the  mechanical  power  expended  upon  the  wheel, 
let,  in  Fig.  177, 

V  =  the  velocity  of  the  water  as  it  enters  the  vanes, 
y  z=  FQR  =  the  angle  between  tiie  tangent  PQ  to 
the  guide  plate,  and  QI^,  the  tangent  to  the 
inner  rim  of  the  wheel, 
Vt  =  the  tangential  component  of  the  velocity, 
Vr=  the  radial  component  of  the  velocity, 
^'j  =:  OQ  —  the  internal  radius  of  the  wheel, 
a   =  angle  between  the  direction  of  the  water  as  it 
escapes  from  the  wheel,  and  XxS' the  tangent 
to  the  outer  rim  of  the  wheel, 
v',  i;/,  v/,  ^2  =  respectively,  the  velocity  of  discharge,  its  tan- 
gential   and    radial   components,    and   the 
radius  of  the  outer  rim. 
Then, 

v^  =  V  cos  /,  Vr  —  v  sin  y  ;  {^^^) 

Vt  =  y'cos  a  ,  Vr'=  y'sin  a  .  (376) 

Tlie  moment  of  the  momentum  of  the  radial  velocities  will 
be  zero ;  hence,  the  moment  of  the  momentum  of  the  water 
as  it  enters  the  wheel  will  be 

Mnvt , 

and  as  it  quits  it,  it  will  be 

hence.  Article  227,  the  work  imparted  to  the  wheel,  neglect- 
ing friction,  contractions,  eddying,  etc.,  will  be 

Pu  —  M{7\Vt  -  r^ol)  00 ;  (377) 


[280.]  TURBINES.  S53 

and  if  h  be  the  head  in  the  supply  chamber  above  the  point  of 
discharge,  the  efficiency  will  be 

e=(^'"' -;»"''>'".  (378) 

Equation  (377)  is  a  fundamental  one  in  the  theory  of  tur- 
bines. The  velocities  v  and  v  will  generally  be  independent 
of  each  other,  depending  upon  the  form  of  the  guide  plates, 
the  vanes,  and  the  speed  of  the  wheel. 

For  the  highest  efficiency,  the  water  must  quit  the  wheel 
with  no  velocity.  There  will,  however,  practically,  always  be 
a  radial  velocity,  but  when  this  is  reduced  to  a  minimum  by 
the  proper  construction  of  the  vanes,  a  maximum  efficiency 
will  be  secured  by  running  the  wheel  at  such  a  speed  as  to 
make  the  tangential  velocity  of  discharge  zero,  in  which  case 
we  have 

v;  =  0, 

Tu  =  Mi\VtGo,  (379) 

-'^.  (380) 

The  coefficients  of  JIf,  equations  (377)  and  (379),  divided  by 
^,  are  the  effective  heads  for  the  respective  cases.  If  to  the 
effective  head  there  be  added  the  head  due  to  the  loss  of 
velocity,  there  being  no  frictional  resistances,  the  sum  will  be 
the  head  in  the  supply  chamber.  The  only  head  lost  in  the 
case  of  maximum  efficiency  will  be  that  due  to  the  radial 
velocity  of  discharge.  For  this  case,  a,  equation  (376),  will  be 
90°,  and  Vr  =  v\  But  if  ^^  be  the  velocity  of  the  water  along 
and  relatively  to  the  vane,  and  Wx  and  w^  the  relative  tangen- 
tial and  radial  velocities,  respectively,  y3  =  the  angle  8XT^ 
Fig.  177,  between  the  tangents  respectively  to  the  outer  rim 
of  the  wheel  and  the  vane  ;  then 

W,-W  cos  ^  ;   Wr^W^lVL  p\\ 

Wr  =  Wt  tan  p ;  \ 

28 


854  TURBINES.  [230.] 

the  head  corresponding  to  which  will  he 

wf  tan^  y^  -^  2^  ; 

hence,  for  the  case  of  maximum  efficiency, 

h  =  (2ri?;,a9  +  wj  tan^  ft)-^2g\  (382) 

from  which  we  find  that  the  velocity  with  which  the  water 
enters  the  wheel  is 

V  =  Vt^QQ  y  =  [— — ■^— ~)  sec  y,       (383) 

Hence  the  velocity  of  entering  the  wheel  varies  inversely  as 
the  velocity  of  the  inner  rim. 

As  limiting  cases  assume  that  ^  =  0,  and  /3  =  0,  aiid  we 
have 

(384) 

v'  =  gh,  (385) 

or  the  velocity  will  be  that  due  to  one-half  the  entire  head. 
If  TiGj  =  ^v,  then 

N  v^  =  2gh,  (386) 

or  the  velocity  will  be  that  due  to  the  head. 
If  TiGD  =  Iv, 

v"  =  4gh, 

or  the  velocity  will  be  that  due  to  twice  the  head,  which  is  an 
hypothetical  condition. 

The  above  analysis  is  equally  applicable  to  the  other  classes 
of  turbines  ;  but  the  construction  must  be  different  for  each 
class  in  order  to  realize  the  conditions  here  imposed.  Thus  it 
will  be  found  that  /3  must  be  larger  for  a  parallel-flow  turbine 
than  foi;  an  outward  flow,  and  still  larger  for  an  inward  flow, 
if  the  transverse  sections  of  the  passages  through  the  wheel 
are  uniform.  Its  value  may  also  depend  upon  Ti  and  /'g?  ^s 
will  be  shown. 


ah 

r^GJ 

If  7\Gi)  =  Vy  then 

[231  •]  TUKBINES.  355 

231.  Special  Cases. — 1.  In  the  Fourneyron  turbine  the 
initial  elements  of  the  guide  plates  arc  radial,  and  passing  out- 
ward from  the  axis  of  the  supply  chamber,  are  gradually 
curved  until  they  finally  terminate  as  nearl}^  tangent  to  the 
inner  rim  of  tlie  wheel  as  practicable,  allowing  sufficient  space 
between  the  plates  for  the  passage  of  the  required  quantity  of 
water.     The  least  value  of  y, 

tan;/  =  ^,  (387) 

gives  the  greatest  possible  tangential  velocity  to  the  water 
when  it  enters  the  wheel. 

The  initial  elements  of  the  vanes  are  also  radial,  but  in 
passing  outward  they  are  curved  backward,  and  at  their 
terminus  make  a  small  angle,  /?,  with  the  outer  rim.  These 
conditions  give 

tan/?  =  ^,  (388) 

In  order  that  the  water  shall  enter  without  impulse,  the 
velocity  of  the  inner  rim  of  the  wheel  must  equal  the  tangen- 
tial velocity  of  the  water,  or, 

Vt  =  7\csD  ;  (389) 

in  which  case  the  initial  velocity  of  the  water  relatively  to 
the  wheel  will  be  the  radial  velocity  Vr.  It  will  then  pass 
along  the  curved  vanes  and  be  finally  discharged  backward 
relatively  to  the  vane  with  a  tangential  velocity  v;^  but  the 
vane  has  a  forward  motion  of  v^gj  =  nrioj ;  hence  the  actual 
tangential  velocity  of  the  water  will  be 

Vt  =  Wt  —  nriCOj  (390) 

which  for  the  best  result  must  be  zero,  for  which  condition 
we  find,  combined  with  equations  (388),  (389),  and  (387), 

Wt  =  nr.GJ  =  -— ^  =  nvt  =  ~~^,  (391) 

tan  p  tan  y  ^  ^       ^ 

.-.  tan  y  =  n  ^  tan  /? ;  (392) 

^  w^ 


366  TURBINES.  t231.] 

which  establishes  the  relation  between  the  outer  angles  of  the 
blades  and  vanes  relatively  to  the  respective  rims  of  the  wheel. 
The  radial  velocity  through  the  wheel  may  be  made  constapt 
when  the  wheel  passages  are  full,  centrifugal  force  being 
neglected,  by  making  the  transverse  sections  of  these  passages 
uniform ;  and  since  their  breadths  vary  as  the  radii  ri  and  r^, 
the  depths  must  vary  inversely  as  the  same  numbers,  thus 
making  the  axial  sections  hyperbolas.  This  being  done, 
Vf.  =  Wr,  and  we  have 

tan  y  =  n  tan  /?,  (393) 

which  is  the  condition  commonly  assumed.  These  conditions 
being  realized,  we  have  for  the  Fourneyron  type — frictional 
losses  being  abstracted — equations  (380),  (382),  (389),  and 
(390), 


v]^ 


^gh         ,  (395) 


2  +  ti"  taii^  fi 
and  from  (379), 

from  which  it  appears  that  the  efficiency  is  greatest  when  ^  is 
least. 

If  yS  =  0 ;  then  ^  =  1  and  v  =  Vgh ;  and  the  head  due  to 
the  velocity  with  which  the  water  enters  the  wheel  is  one-half 
that  due  to  the  head  in  the  supply  chamber.  Since  the  water 
enters  the  wheel  with  a  small  velocity  relatively  to  the  vanes, 
Vt  tan  y,  and  is  discharged  with  a  greater  relative  velocity, 
w  =  no^GJ  sec  /?,  it  follows  that  from  the  point  of  entrance  the 
velocity  of  the  water  relatively  to  the  vane  is  accelerated  to 
the  point  of  discharge,  while  the  actual  velocity  diminishes 
along  the  same  path.    * 

2.  In  the  cup  vane  turbine  the  guide  plates  are  essentially  the 
same  as  the  Fourneyron  type ;  but  the  water  issues  from  them 


[231.]  TURBINES.  357 

with  a  velocity  due  to  the  head  in  the  supply  chamber,  and 
hence  enters  the  wheel  as  a  jet.  The  vanes  are  vertical,  and 
their  surfaces  are  as  nearly  semi-cylindrical  as  the  circum- 
stances of  construction  will  permit.  To  avoid  loss  from  shock, 
the  initial  elements  of  the  vanes  should  be  tangential  to  the 
direction  of  the  water  relatively  to  the  vanes  as  it  enters  them. 
The  velocity  of  the  inner  rim  of  the  wheel  should,  according 
to  Article  (211),  equal  one-half  the  tangential  velocity  of  the 
water  as  it  enters  the  vanes ;  hence 

TiCD  =  Jv  cos  y,  (397) 

The  radial  velocity  of  the  water  as  it  quits  the  guide-plates 
will  be 

Vr  =  v  sin  y  ;  (398) 

hence,  if  y'  be  the  angle  between  the  vanes  and  the  inner  rim 
of  the  wheel  at  their  intersection,  we  have 

tan  y'  =  ZillLil  =  2  tan  y.  (399) 

'        \v  cos  y  ^ 

If  y  =  0,  y'  will  also  be  zero,  or  the  terminal  elements  of 
the  guide  plates  and  the  initial  elements  of  the  vanes  will 
both  be  tangent  to  the  rim  of  the  wheel.  The  radial  velocity 
of  the  water  being  unaffected  by  the  motion  of  the  wheel,  and 
the  tangential  velocity  relatively  to  the  wheel  being  one-half 
that  due  to  the  head,  the  velocity  relatively  to  the  vane  will  be 


v  =  Jv\/cos^  y  +  ^  sin^  y,  (400) 

and  if  the  motion  were  rectilinear,  the  velocity  relatively  to 
the  vane  would  be  uniform,  and  the  water  would  quit  the 
wheel  with  a  velocity  v'  =  w  \  but  since  the  velocity  of  the 
elements  of  the  vanes  varies  directly  as  their  distance  from  the 
centre  of  the  wheel,  the  relative  velocity  will  vary.  But,  in 
any  case,  we  have,  for  best  efficiency, 

nr^^oo  =  w  cos  /?  ;  (4^1) 

and  the  velocity  lost  will  be 

w  sin  ft. 


358  TUEBINES.  [231.] 

The  energy  imparted  to  tlie  wheel  will  be,  equations  (379), 
(397), 

Pu  =  Mvt'Ti  CO  =z  Mv  COS  Y'^v  COS  ;/, 

=  Wi^  cos^  y.  (402) 

The  kinetic  energy  of  the  water  as  it  leaves  the  gates  will 
be,  W  being  the  weight  of  water  flowing  per  second, 

Wh  =  iMv" ;  (403) 

hence  the  theoretical  efficiency  will  be 

(404) 


iif^  C0s2  y 


3.  In  the  inward  flow  turbine  the  initial  elements  of  the 
guide  curve  are  at  the  outer  circumference  of  the  supply 
chamber,  and  curve  so  as  to  form  a  small  angle  with  the  outer 
rim  of  the  wheel.  The  outer  elements  of  the  vanes  are  radial, 
and  as  they  approach  the  inner  rim  of  the  wheel,  they  curve 
in  a  direction  opposite  to  that  of  the  guide  plates.  Since  n^  in 
equation  (393),  will  be  fractional,  the  angle  &  will  exceed  y. 
The  water  should  enter  the  wheel  with  a  tangential  velocity 
equal  to  the  velocity  of  the  outer  rim.  The  remainder  of  the 
analysis  is  the  same  as  for  the  Fourneyron  turbine.  If  the 
vanes  are  cup-shaped,  the  analysis  will  be  the  same  as  the 
second  case  above. 

4.  In  the  parallel  flow  turbine,  the  initial  elements  of  the 

guide  plates  are  vertical,  and  are  curved  as 
they  pass  downward  so  as  to  terminate  as 
nearly  horizontal  as  possible.  The  initial 
elements  of  the  vanes  are  usually  vertical, 
and  the  vanes  curve  in  an  opposite  direc- 
tion from  that  of  the  guide  plates,  termi- 
nating at  the  same  angle  with  the  horizontal. 
This  last  is  a  result  of  making  n  =  1  in 
equation  (393),  giving 

y  =  ft. 


[232.]  TURBINES.  869 

5.  If,  in  an  out/ward  flow  turbine,  the  vanes  make  an  angle 
with  the  inner  rim,  exceeding  90°  and  less  than  180°,  which 
will  be  an  angle  greater  than  for  the  Fourneyron  type,  and 
less  than  for  the  cup  vane,  we  shall  find 

TOO  >  ^/gh  and  <  "^^gh^ 

for  a  speed  of  maximum  efficiency. 

6.  An  unlimited  number  of  turbines  having  different  forms 
of  vanes  may  be  devised  which  will  give,  theoretically,  high 
efficiencies ;  but  if  friction,  whirls,  contractions,  etc.,  be  in- 
volved, those  forms  will  be,  practically,  most  efficient,  which 
for  the  same  theoretical  efficiency,  have  the  least  of  the  above 
named  resistances. 

232.  General  case. — In  any  hydraulic  motor,  let  H  be 
the  total  available  head,  h  the  loss  of  head  due  to  location  and 
form  of  wheel,  or  that  portion  of  the  head  below  the  wheel 
(and  possibly  above)  not  utilized,  h!  the  loss  by  shock,  h"  the 
head  lost  on  account  of  the  velocity  of  the  water  at  quitting 
the  wheel,  h'"  the  head  lost  from  contractions,  whirls,  friction, 
etc.,  then  will  the  mechanical  power  be 

Q^Q[H-h-h'  -  h"  -  A'"] ;  (405) 

and  the  gross  efficiency  of  the  motor  will  be 

e  =  ^ .  (406) 

The  efficiency  of  the  motor  as  a  machine  may  exceed  this, 
for  it  may  be  so  constructed  and  placed  that  its  entire  efficiency 
will  be  developed  with  a  less  head  than  H;  as,  for  instance, 
in  the  case  of  an  overshot  wheel,  if  there  were  4  or  5  feet  fall 
below  the  lowest  point  of  the  wheel,  it  would  in  no  way 
affect  the  efficiency  of  the  wheel,  but  the  wheel  would  fail  to 
utilize  the  power  of  the  water.  In  determining  the  quality 
of  the  wheel  as  a  motor,  only  so  much  of  the  head  should  be 


860  TUKBINES.  [232.] 

considered  as  is  actually  necessary  for  operating  the  wheel  in 
the  place  where  it  is  established. 

EXAMPLES. 

1.  A  stream  delivers  10,000  gallons  of  water  per  minute  to 
a  Foiirnejroii  turbine.  If  the  fall  be  100  ft.,  tan  /3=  ^^n=  1.4, 
and  revolutions  per  minute  =  240 ;  lind  the  internal  and 
external  radii  for  the  most  efficient  speed,  the  HP,  and  the 
depth  of  the  casing  at  the  outer  periphery. 

From  equation  (395)  we  find 


i/i^ 


,  ,  -  ,.  324-  X  100 


(1.4)2  X  i  ' 

or 

n  =  2.14  feet. 

r^  =  m\  =  2.996  feet. 
From  (394), 

efficiency  =  ^^-j^^^^, 

=  .90. 

^    ^  10,000  X  231  X  62i  x  100  x  .9 
~  1,728  X  33,000  * 

=  227.8. 
From  (391), 

Wr  =  Vr  =  nriGD  tan  ft, 
or 

Vr  =  25.1328  feet  per  second, 
and 

,__   10,000  X  231  X  12 
^  ■"  1,728"  X  25.133  x  6;r  x  60 

=  .66  inches. 

2.  An  outward  flow  turbine  is  supplied  with  water  having 
a  head  of  350  feet.  The  internal  diameter  is  15  inches,  and 
the  external  diameter  is  21  inches.     Angle  of  guiding  plates^ 


[232.]  TURBINES.  861 

;/  =  16°,  and  the  vanes  of  the  wheel  are  placed  at  the  corre- 
sponding angle.  Depth  of  wheel  at  the  inner  periphery, 
2  inches.  Find  the  number  of  revolutions  per  minute  required 
for  the  best  efficiency,  neglecting  frictional  resistances. 

Taking  account  of  frictional  resistances,  and  supposing  the 
true  efficiency  to  be  fths,  find  the  best  number  of  revolutions 
and  HP  developed. 

From  equation  (395), 


v,=^ 


=  104  feet  per  second. 


2.0822 

104  X  60 

2;r  X  f 


=  1,589  revolutions  per  minute. 


If  we  take  frictional  resistances  into  account,  and  suppose 
the  wheel  to  be  running  at  its  greatest  efficiency,  we  will  have 
for  the  total  energy, 

mgQh  =  m  Qvi  +  ^  v^n^  tan^  /^  +  ^  2Fv't  (1  +  n'  tan^  /?). 

Therefore  the  efficiency  is 

2 


e  = 


or 


2  +  n^  tan^  /3  +  2F{1  +  rv'  tan^  p) ' 

v^  =  Ighj     or    Vt  =  91.9  feet  per  second, 

91.9  X  60       ,   ..^         ,  ^. 
Wi=  — r —  =  1,440  revolutions  per  mmute, 

Wf.  =  Vr  =  Vt  tan  y, 
Wf  =  26.35  feet  per  second. 
Area  of  flow  at  inner  periphery  =  .6545  square  feet. 
.6545  X  26.35  x  350  x  62.5  x  60  x  .Y5 


J2P  = 


33,000 
=  514.45. 


362  BESISTANCES.  [233.] 

3.  Which  requires  the  larger  gate  opening  for  tlje  same 
work,  the  Fourneyron  turbine  of  Case  1,  or  the  cup  vane  tur- 
bine of  Case  2  ? 

4.  If  the  head  in  the  supply  chamber  be  15  feet,  find  the 
velocity  of  the  inner  rim  for  best  efficiency  in  Cases  1  and  2, 
when  ^  =  0,  and  when  /J  =  20°. 

5.  Explain  why  the  same  quantity  of  water  flowing  through 
the  cup  vane  as  through  the  Fourneyron  turbine  produces  the 
same  work,  when  the  velocities  of  the  water  on  entering  the 
wheel  and  also  the^velocities  of  the  wheel  are  different. 


KESISTANCES. 

233.  Sudden  enlargements. — If  a  stream  flows  along  a 
pipe  liaving  a  sudden  enlargement,  it  suffers  a  loss  of  velocity. 
If  V  be  the  velocity  in  the  small  pipe,  and 
^'2  that  in  the  enlarged  part,  it  might  seem 
that  the  head  lost  would  be  (v^  —  v^^  -^  2^ ; 
but  this  is  true  only  when  the  internal 
forces  are  functions  of  the  distance  between 
the  centres  of  force,  Article  (151),  a  condi- 
tion not  realized  in  this  case.  The  particles  act  upon  each 
other  according  to  a  law  not  definitely  known.  It  may  be 
partly  of  the  nature  of  friction,  but  chiefly  it  results  in  the 
production  of  whirls  at  the  angles  of  the  tube.  It  is  also 
known  that  the  reduction  of  the  kinetic  energy  develops  heat, 
and  this,  being  retained  in  the  tube,  partially  maintains  the 
head,  so  that  the  loss  is  not  as  great  as  it  otherwise  would  be. 
The  action  is  not  exactly  of  the  nature  of  the  impact  of  finite 
bodies,  but  resembles  it  in  regard  to  the  non-elasticity  of  the 
water  and  the  comparatively  sudden  reduction  of  the  velocity ; 
and  for  these  reasons  we  analyze  the  case  as  for  inelastic  im- 
pact. In  all  cases,  however,  when  the  law  of  action  is 
unknown,  the  result  should  be  checked  by  direct  experiment. 
We  may  assume  that  the  mass  m  passing  out  of  the  small  tube 
in  a  unit  of  time  impinges  upon  the  mass  M  in  the  enlarged 


Fig.  179. 


[234.  J  RESISTANCES.  363 

part  of  the  tube ;  then,  according  to  equation  (42),  the  loss  of 
energy  will  be 

imiv-v^y,  (407) 

and  it  is  found  that  this  expression  represents  with  sufficient 
accuracy  the  results  of  experiment. 

This  amount  of  energy  transformed  into  heat  would  raise  a 
pound  of  water  a  nu giber  of  degrees  in  temperature  given  by 
the  expression 

_  m{v-v,y  . 

^  -    2  X  772    '  ^^^^J 


where  772  is  Joule's  mechanical  equivalent  of  heat. 
The  head  lost  will  be 

(jLpl.  (409) 


234.  Resistances  in  long  pipes. — The  principles  of  work 
and  energy  on  which  Bernoulli's  theorem  is  founded  may  be 
extended  to  the  case  in  which  there  is  resistance  along  the 
stream  between  A  and  B,  as  is  the  case  in  actual  tubes.  The 
law  of  the  resistance  can  be  determined  only  by  experiment. 
It  has  been  found  that,  with  sufficient  accuracy,  the  resistance 
varies  as  the  perimeter  of  contact  between  the  liquid  and  pipe 
in  a  transverse  section,  called  the  wetted perirneter,  also  as  the 
square  of  the  velocity,  and  a  factor/  dependent  upon  the  con- 
dition of  the  pipe.  Hence,  if  B  be  the  resistance  at  any  sec- 
tion, *,  w  the  weight  of  a  unit  of  volume  of  the  fluid,  and  c 
the  wetted  perimeter,  we  have 

-^       2g 

If  the  velocity  and  sections  be  uniform,  then  will  the  loss 
of  head  for  a  length  I  be 


864  RESISTANCES.  [385.] 

The  value  of/  for  rivers,  as  given  by  Ey telwein,  is 


where 


/=«  +  -, 


a  =  0.007164,    h  =  0.007409. 


(411) 


235.  G-ENEKAL  CASE. — Let  AB  be  a  stream  flowing  from  a 
reservoir  wliose  upper  surface  is  at  C,  and  whose  exit  is  at  B, 
Or,  to  generalize  it,  let  B  be  any  point  in  the  pipe.  Through 
B  draw  a  horizontal  GB,  then  will  OG  =  BD  be  the  total 


Fig.  180. 

head  on  B,  neglecting  the  pressure  of  the  air  as  we  may  do  in 
most  practical  cases.  If  B  be  the  exit,  let  BL  be  the  head 
necessary  to  overcome  the  pressure  at  the  orifice.  If  the 
opening  be  the  full  size  of  the  pipe,  BZ  will  be  zero  ;  but  if 
it  be  contracted,  it  will  be  of  finite  value.  Lay  oft*  on  BD, 
the  following  heads : 


DE 
HI 


h 


2^ 


the  head  due  to  the  velocity  of  exit, 


z=  h^  =  the  head  lost  due  to  the  resistance  of  entering 
the  pipe  at  A, 
IM  =  /iq  =  the  head  necessary  to  overcome  the  resistances 

along  the  straight  portions  of  the  pipe, 
MN=  ^4  =  head  lost  by  bends  in  the  pipe, 
NO  =z  h^  =  head  lost  by  sharp  angles. 


[235.]  ^  RESISTANCES.  865 

OL   =  /(q  —  head  lost  by  enlargements  and  contractions, 
LB  =  h-  =  head  due  to  pressure  at  B, 
DB  =  H=  total  head  at  B. 

Then 

H^K-V  lu  +  h  +  ^4  +  ^5  +  h  +  K      (412) 

The  values  of  these  several  heads,  with  the  exception  of  the 
first,  are  determined  by  experiment.  "  They  are  usually  de- 

termined  as  multiples  of  ^^  =  —  ,  and  hence  may  be  written 

with  the  use  of  corresponding  subscripts,  thus 

H=  (1  +/,  +/,  +/,  +/,  +/«  +/,)|.     (413) 

The  following  are   some  of  the   principal  values  of   the 
coefficients  as  determined  by  experiment : 

Orifice  in  a  thin  plate,  which  would  be  applicable  when  the 
length  of  the  pipe  is  zero, 

/,  =  0.054. 

Straight,  short  cylindrical  mouth-piece  perpendicular  to  the 
side  of  the  vessel, 

/a  =  0.505. 

For  the  resistance  along  the  pipe  whose  length  is  ?,  section 
«,  wetted  perimeter  c, 

Where/*  is  dependent  upon  the  dimensions  of  the  pipe,  and 
according  to  Darcy,  its  value  is 


/=.(i  +  i). 


366 


KESISTANCES. 


[235.] 


Where  d  is  the  internal  diameter  of  tlie  pipe  in  inches,  and 
Ic  for  clean  cast-iron  pipe,  is 

A;  =  0.005; 
and  for  pipe  very  rough  from  sediment, 

Tc  =  0.01. 
According  to  Weisbach,  for  clean  cast-iron  pipe, 

0.0043 


/  =  0.0036  + 


^/v 


where  ^  is  the  mean  velocity.     The  larger  computed  value  of 
f  would  ordinarily  be  used. 

For  a  bend  through  an  angle  ^,  the  radius 
of  the  curve  being  r^  diameter  of  the  pipe 

i|0.131. 1.847©*}. 


Fig.  181. 


/4 


For  a  knee,  or  sharp  angle,  i^ 

/s  =  0.946  sin^  |  +  2.05  sin^  \r  , 

and  if  the  knee  is  90°,  this  becomes  ^ 

/b  =  1  nearly. 

The  value  of  f^  may  be  computed  approximately,  and  f^j  is 
de])euclent  upon  the  condition  of  the  opening  at  exit. 

Tliese  values  in  (413)  enable  one  to  determine  v\  ov  v  and 
the  other  quatitities  being  known,  H,  the  necessary  head,  may 
be  determined.  Indeed  any  one  of  the  many  quantities 
may  be  found  in  terms  of  the  others. 

To  find  the  pressure  on  the  pipe  at  any  point,  as  J,  lay 
down  from  CD  the  head  dh  due  to  the  .velocity  at  h,  hi  the 
head  lost  at  entrance  A^  im  the  head  lost  by  the  resistance  of 
the  straight  part  of  the  pipe  from  A  to  Z>,  7nn  the  head  lost  by 


[235.]  RESISTANCES.  367 

bending  between  A  and  h  due  to  bends,  no  that  lost  by  angles, 
ol  all  other  heads  lost ;  then  will 

lb  =  pressure  on  pipe  at  h. 

The  head  hg  is  potential  in  reference  to  BG,  The  line  T^Z 
limits  the  upper  ends  of  the  heads  due  to  pressure  on  the 
pipe  ;  and  if  the  pipe  lay  along  this  line  there  would  be  no 
pressure  except  on  the  under  side  due  to  the  weight  of  the 
fluid.  If  an  orifice  be  made  at  5,  the  liquid  would  flow  out 
in  a  stream  rising  to  the  height  hi,  less  the  resistance  at  exit 
from  said  orifice  and  the  resistance  of  the  air.  But  if  the  pipe 
lay  along  FZ,  the  liquid  would  not  flow  out  of  an  orifice  at  I. 

E  XAMPLES. 

1.  Let  it  be  required  to  deliver  water  with  a  given  hydraulic 
head  //q?  such,  for  instance,  as  is  necessary  to  drive  an  engine. 
Then 

2g  2g       '^  d2g' 

where  22^ \a  the  sum  of  the  coefficients  due  to  valves,  bends, 
knees,  sharp-edged  entrance,  etc.,  and  /  that  for  surface  fric- 
tion.    Therefore  we  have 


^^      l  +  ^i^+4/i' 

from  which  the  velocity  is  obtained. 

If  Q  be  the  required  quantity,  and  S  the  section,  then 

2.  Water  flows  from  a  tank  through  a  1  in.  vertical  pipe. 
Find  the  head  in  the  tank  so  that  the  velocity  of  discharge 
may  be  the  same  for  every  length,  taking  into  account  the 
resistance  of  sharp-edged  entrance  as  well  as  of  surface  fric- 
tion. 


368  RESISTANCES.  [235.J 

Let  I  be  the  length  of  the  pipe,  and  H  the  head  above  the 
orifice ;  then 

or 


^=<'-/.'£-(£#-'> 


2^  '    \^g8 

But  the  conditions  require  that  the  term  containing  I  shall 
disappear ;  hence 

and 

where  v  is  independent  of  I  as  required. 
Then 


and 


^  (1  +  i) 


x.005(l+i) 


4 
=  3.7  feet. 

3.  Water  flows  from  a  tank  through  a  uniform  sloping  pipe 
of  diameter  d.  Taking  into  account  the  resistance  at  entrance, 
show  that  the  water  will  flow  with  the  same  velocity  what- 
ever be  the  length  of  pipe,  if  the  pipe  slopes  at  an  angle, 

sin  ^  =  ^  r  --t,"  )  ^^  ^0  being  the  head  in  the  tank  above 

the  entrance. 


L236.J  GASES.  369 

We  have,  as  before, 

We  will  supi^ose  6  so  small  that  the  resistance  to  sharp- 
edged  en  trance  may  be  taken  as  F=  i.  We  have/*=:  .005  — -r— , 

and  m  =  -r  '     Hence  we  have 
4 

which  may  be  regarded  as  an  equation  of  identity  in  I,   There- 
fore equating  like  terms, 


or 


Also 


sm  ^  =  ^  ^ 
2^   d 


.04(^_+J)^ 


GASES. 

236.  The  density  of  an  ideal,  incompressible  fluid  is  inde- 
pendent of  the  pressure  to  which  it  is  subjected,  and  depend- 
ent only  upon  its  constitution  ;  but  the  density  of  a  compress- 
ible fluid  is  dependent  upon  both  its  constitution  and  external 
pressure.  Thus,  if  water  were  strictly  non-compressible,  its 
density  at  all  depths  in  the  ocean  would  be  the  same  as  at  the 
surface;  but  the  atmosphere,  being  a  gas,  diminishes  in  den- 
sity as  we  ascend,  since  the  weight  of  the  atmosphere  above 
24 


870  GASES.  [237.] 

any  point  is  less  the  higher  the  point.  The  laws  of  the 
pressure  of  fluids  given  in  Articles  174  to  178  inclusive  are 
applicable  to  gases.  But  if  the  pressure  varies,  the  density 
and  temperature  both  vary  according  to  laws  which  are  deter- 
mined by  experiment. 

237.  Boyle's  (or  Mariotte's)  Law. — According  to  tiie 
experiments  of  Boyle  and  Mai-iotte,  the  volume  of  a  gas  at 
uniform  temperature  varies  inversely  as  the  pressure  to  which 
it  is  subjected.  Hence,  if  v^  be  a  known  initial  volume  of 
gas,  and^o  the  initial  pressure  per  unit  of  area  to  which  the 
gas  is  subjected — being  the  pressure  upon  its  bounding  sur- 
face, or  at  any  point  within  it,  for  which  latter  reason  it  is 
also  called  the  tension  of  the  gas — and  v  and  p  any  other  con- 
temporaneous volume  and  pressure  ;  then  we  have 

pv  =-PqVq  =  constant  =  m,  (414) 

which,  if  p  and  v  vary  simultaneously,  is  the  equation  of  an 

equilateral  hyperbola  referred 

to  its  asymptotes.     In  testing 

this   law  experimentally,  it  is 

necessary,  after  compressing  or 

dilating  the  gas,  to  make  the 

temperature   of    the    gas    the 

^^-B___   same  as  the  initial  temperature  ; 

>       i^        in  the  fornier  case  cooling,  and 

I         in  tlie  latter  heating  it.     The 

0    X    law  is  found  to  be  very  nearly, 

^  but  not  exactly,  correct  for  air 

and  other  known  gases,  the 
agreement  being  nearer  the 
more  perfect  the  gas,  from  which  it  is  inferred  that  it  cor- 
rectly represents  the  law  for  the  ideally  perfect  gas. 

Since  the  density,  S,  varies  inversely  as  the  volume,  we 
have,  for  uniform  temperature, 

6v  =  SqVq  =  constant,  (415) 

and 

dp,:^6op.  (416) 


Fig.  182. 


[388.]  GASES.  371 

238.    To   FIND  THE  TENSION   AT   ANT   POINT   OF   A  COLUMN  OF 

GAS  OF  UNIFORM  TEMPERATURE. — On  accouiit  of  the  Weight  of 
the  gas  and  its  compressibility,  its  density  will  vary  as  some 
function  of  the  height.  Conceive  a  prismatic  column  of  gas 
whose  base  is  unity,  and  at  the  lower  base  p^  the  tension,  S^ 
the  density,  w^  the  weight  of  a  unit  of  volume,  and  at  height 
z  let  the  corresponding  quantities  be  ^,  (J,  w.  Then  will  the 
pressure  on  the  lower  base  equal  the  pressure  at  the  height  z 
added  to  the  weight  of  the  prism  of  gas  of  that  height ;  or, 

and  differentiating, 

dp  =  —  wdz^  (417) 

which  is  independent  of  the  law  of  pressure  and  of  the  total 
head. 
According  to  Mariotte's  law, 

S:S,::'^:'^^::p:j,„  (418) 


and  considering  gravity  as  uniform. 


w  =^V  (419) 


Substituting  and  reducing. 


integrating  between  the  limits  j9  and^o,  z  and  0,  gives 
log£-=-!f!o 


\p=p^    ^«    .  (420) 


372                                                   GASES. 

[239,  24( 

Similarly, 

(421) 

The  weight  of  a  prism  of  gas  of  height  A,  will  be 

W  =  \wd3  —  Wq     e    Pu    dz  :=  Po   1  —  e    po 

(422) 

If  A  =  CO ,  TF  =  j?05  as  it  should. 

239.  N^UMERicAL  VALUES.  —  The  mean  pressure  of  the 
atmosphere  at  the  level  of  the  sea  is  po  =  14.7  pounds  per 
square  inch,  or  2116.8  pounds  per  square  foot. 

The  weight  of  a  cubic  foot  of  pure  dry  air  under  the  press- 
ure of  14.7  pounds  per  square  inch,  and  at  the  temperature  of 
melting  ice  (32°  F.), 

Wq  =  0. 0S07 2S  _pmmd  avoirdupois.  (423) 

If  the  atmosphere  were  pure,  dry,  and  uniform,  and  of  the 
density  as  at  present  at  the  level  of  the  sea  where  the  tem- 
perature is  32°  F.,  the  height  would  be 

"^=^^8= '''''' '^''  (*^*) 

or  the  height  of  one  atmosphere  of  uniform  density  is  nearly 
5  miles;  but  as  the  atmosphere  does  not  fulfil  these  condi- 
tions, and  being  a  little  heavier,  the  height  is  a  little  less  than 
this  value. 

At  the  level  of  the  sea  the  barometer  stands  at  29.92  inches  nearly. 


At  5,000  feet  above  '' 

24.7 

At  10,000  feet  (Mt.  ^tna) 

20.5 

At  15,000  feet  (Mt.  Blanc) 

16.9 

At  3  miles                                  " 

16.4 

At  5  miles                                  *' 

8.9 

240.  To  find  the  tension  at  extreme   heights  when  the 
variation  of  gravity  is  considered. 


.]  GASES.  ^73 

From  equation  (418)  we  have 

«,  =  ^.|^.  (425) 

Let  Id  be  the  radius  of  the  earth,  s  any  distance  above  the 
earth,  then 

and  (417),  (425),  (425a),  give 

^  ^P  —       '^0       ^       ^ 

integrating  between  the  limits  oip  and^o>  ^  and  0, 
1     P  ^  _'^^    ^^ 


and  similarly, 


p=.p,e    ^-^^*; 


_  Wq    Bz 

S=6,e'  ^^''''*  (425 J) 


Assuming  the  density  of  air  at  the  earth  Sq  =  -^  of  a  pound 
per  cubic  foot,  as  it  is  very  nearly,  and  R  =  20,860,000  feet, 
and  equation  (425J)  reduces  to 


+  z 


<J  =  tU10)       "    %  (425c^) 

If  ;5  =  CO ,  we  have 

1 


400  X  (10)5 


(425cZ) 


which  18  the  limit  of  the  density.     If  gravity  be  considered 
uniform,  we  would  have  for  the  height  equal  to  the  radius  of 

*  A  more  cumbersome  demonstration  of  this  formula  may  be  made  by 
means  of  Prop,  xxii.,  B.  II.,  of  Newton's  Principia, 


374  GASES.  [241.] 

the  earth,  z  =  20,860,000,  and  (421)  and  (418)  reduce  to  the 
same  as  (425d). 

241.  Heights  may  be  determined  approxiinately  by  the 
pressure  of  the  atmospliere.  If  its  temperature  were  uniform 
and  free  from  currents,  we  would  have  from  equation  (420), 

,  =  ^«logB=  26,22110-.^% 

where  Jq  ^^^  ^  ^I'e  the  readings  of  a  barometer  at  the  lower' 
and  higher  stations  respectively.     Adapting  this  to  the  use  of 
common  logarithms,  we  have 

9Q  Q9 
z  =  26,221  X  2.30258  log  .^^ 

=  60,73d  log— ^  . 

But  this  coefficient  is  known  to  be  too  large,  and  practice 
sliows  that 

9Q  Q9. 
2  =  60,345  log  ^^  (426) 

gives  better  results. 

If  Zi  be  the  height  of  another  station,  then 

Zi  =  60,345  log  — 7 —  ; 
fh 

from  which  subtracting  the  former,  we  have 

z,-z  =  h  =  60,345  log  I .  (427) 

This  result,  however,  requires  to  be  corrected  for  the  effect 
of  the  temperature  on  the  mercurial  column,  the  effect  of  the 
variation  of  gravity  due  to  great  heights,  change  of  gravity 
due  to  latitude,  the  reduction  of  observations  when  not  simul- 
taneous, tlie  effect  upon  the  mercurial  column  due  to  the 


[242.]  GASES.  375 

attraction  of  the  mountain  on  which  observations  are  made, 
all  of  which  are  well  discussed  by  Poisson  in  his  Traite  de 
Mecanique, 

li  T  =  the  temp'ture  of  attached  thermometer  at  lower  station, 


7" u                    (( 

it 

upper     " 

t     =             "               detached 

a 

lower     " 

t'    =             "                    '' 

a 

upper     " 

^  =  the  latitude  of  the  place ; 

then,               } 

A  -  60,345  ^^^••'''^•'^(^ 

+  t'- 

■64) 

'            1  -  0.002551 

[  COS  5 

Icp 

"" '°«-  U  ■  L  + 0.0001001  (r-nJ  *''*•    ^^^^'^^ 

Observations  should  be  taken  at  both  stations  simultane- 
ously, in  calm  weather ;  but  when  this  cannot  be  done,  two 
observations  may  be  made  at  one  station  at  different  hours, 
and  one  at  the  other  at  a  time  midway  between,  and  the  mean 
of  the  former  be  used  as  one  observation. 

242.  GrAT  LussAc's  (or  Charles')  Law. — It  was  found  by 
these  men  by  direct  experiment,  that  the  increase  of  tension 
in  a  fixed  volume  of  gas  is  directly  proportional  to  the  increase 
of  temperature. 

If^  Iq,  -yoj^®  respectively  the  initial  pressure,  temperature, 
and  volume  of  a  gas,  and  p,  U  'V,  any  other  corresponding 
contemporaneous  values,  and  /?  a  coeflBcient  of  expansion, 
then  we  have,  the  pressure  being  constant, 

v  =  v,{l  +  fi  {(-(,)),  (428) 

the  volume  being  constant, 

p  =  M'i-+^i*-to)).  (429) 

From  equation  (428), 


376  GASES.  [243.] 

In  practice,  the  initial  temperature  is  taken  as  that  of  melt- 
ing ice,  32°  F.,  or  0°  C,  and  the  initial  pressure  that  of  one 
atmosphere,  or  147  lbs.  per  square  inch,  or  29.92  inches  of 
mercury  (760  millimeters).  The  initial  volume  may  be  of 
any  convenient  value.  For  these  values  it  has  been  found 
that 

for  carbonic  acid,  ^  =  0.0037099  for  1°  C, 

for  air,  ^  =  0.0036706 

for  hydrogen,         /5  =  0.0036613 

From  these  values  it  appears  that  the  more  perfect  the  gas, 
the  less  will  be  the  coefficient  of  expansion ;  and  Kankine 
concluded,  in  accordance  with  the  theory  of  molecular  vortices, 
that  the  limiting  value  is 

^  =  .000365  =  ^j  for  1°  C. 

=  0.0020275  =  — 1^  for  1°  F.,  (431) 

which  values  are  now  used  for  the  ideally  jperfect  qas^  and  are 
sufficiently  accurate  for  air  and  several  other  of  the  more 
perfect  gases. 

243.  Absolute  zero.  —  Equation  (429)  becomes,  for 
^o  =  0°C., 


i>=i>o(l+2fe),  (432) 


in  which,  if  ^  =  —  274°,  jp  becomes  zero,  and  a  perfect  gas 
would  be  destitute  of  tension.  Similarly,  according  to  (428), 
the  volume  would  vanish.  In  Fahrenheit's  scale  t^  is  32°, 
and  equation  (429)  becomes 

t  -  32 


^=^1-^-1^1)'  (*^.^) 


in  which,  if  ^  —  —  461.2,  p  becomes  zero  as  before.  Accord- 
ing to  the  modern  theory  of  heat,  at  this  temperature  the 
molecules  of  the  gas  would  be  at  perfect  rest.  It  is  the  point 
of  absolute  deprivation  of  heat,  and  is  called  the  zero  of  abso- 


[244.]  GASES.  377 

lute  tem;perature.  No  such  temperature  can  even  be  approxi- 
mated by  any  known  process,  the  lowest  recorded  temperature 
obtained  by  artificial  means  being  —140"  C.  (—284°  F.) ; 
but  an  extension  of  the  law  of  uniform  expansion  leads  to 
this  result.  Temperatures  reckoned  from  the  absolute  zero 
are  called  ahsohite  temperatures^  and  are  especially  useful  in 
simplifying  many  formulas  in  regard  to  heat.  If  t  be  the 
temperature  on  Fahrenheit's  scale,  and  r  the  same  tempera- 
ture on  the  aJbsolute  scale,  Tq  =  493.2,  the  temperature  of 
melting  ice,  then 

r=46r.2  +  i{=ro-32^  +  t, 
and  equation  (433)  becomes 

P  =i'07  •  (434) 


If  w  be  the  weight  of  a  cubic  foot  of  dry  air  at  pressure  p 
pounds  per  square  inch,  and  at  absolute  temperature  r,  then 

'^  =  '«o|;f;  (435) 

and  if  w'  be  the  weight  at  temperature  r'  and  tension  jt?',  then 

^'=^^1;     ^^=(^0^.-^  (436) 

p    r'  p^     r'  ^       ' 

and  since  for  gravity  uniform,  the  volume  varies  inversely  as 
the  weight, 

v  =  ro=^«.I.  (437) 

Since  the  mechanical  properties  of  gases,  whether  at  rest  or 
in  motion,  involve  the  property  of  heat,  we  now  consider 
some  of  the  abstract  properties  of  the  latter. 

244.  Heat  is  a  fomi  of  energy.  It  is  not  force,  since 
force  is  onl^*^  one  of  the  elements  producing  energy.  See 
Articles  25,  26,  151.     It  is  believed  to  be  a  certain  manifesta- 


378  GASES.  [245,  246.] 

tion  of  the  motion  of  the  particles  of  a  body.*  The  following 
principles,  the  result  of  observation  and  experience,  have 
become  established. 

1.  Heat  may  be  transformed  into  external  work ;  and  con- 
versely, work,  under  certain  conditions,  may  be  transformed 
into  heat. 

2.  Heat  cannot  be  transferred  from  a  body  of  a  lower  to 
one  of  a  higher  temperature  except  by  the  aid  of  a  machine 
and  the  expenditure  of  mechanical  energy. 

245.  The  thermal  unit  is  the  .amount  of  heat  energy 
necessary  to  raise  a  unit  of  weight  of  ice-cold  water  one  degree 
on  the  thermometric  scale.  The  English  thermal  unit  is  the 
amount  of  heat  energy  necessary  to  raise  one  pound  of  water 
from  the  temperature  of  °32  F.  to  33°  F. ;  the  value  of  which, 
as  found  by  Joule,  is  the  heat  produced  by  friction  in  bringing 
a  body  weighing  one  pound  to  rest  after  falling  772  feet  in  a 
vacuum.f  Or,  more  briefly,  the  English  thermal  unit  equals 
772  foot-pounds  of  work.  This  is  called  Joule's  equivalent, 
and  is  usually  represented  by  J. 

The  equivalent  in  French  units  is  the  heat  energy  neces- 
sary to  raise  one  kilogramme  of  water  from  0°  C.  to  1°  C, 
and  equals  424  kilogramme-metres. 

The  amount  of  heat  energy  necessary  to  raise  the  tempera- 
ture 1°  is  not  the  same  from  all  temperatures,  although  for 
ordinary  mechanical  purposes  it  is  considered  constant  for  the 
same  substance. 

246.  Specific  heat  is  the  heat  necessary  to  raise  a  unit  of 
weight  of  any  substance  one  degree  on  the  thermometric 
scale,  the  thermal  unit  being  unity. 

Strictly  speaking,  specific  heat  is  a  ratio,  being  the  quotient 
obtained  by  dividing  the  quantity  of  heat  required  to  raise 
the  temperature  of  one  pound  of  the  substance  one  degree  by 
the  quantity  required  to  raise  the  temperature  of  one  pound 
of  water  the  same  amount. 

*  Elementary  Mechanica,  p.  69.     Stewart  on  Heat. 
\  Elementary  Mechanics, -pp.  HI,  72. 


[247,  248;]  GASES.  379 

The  specific  heat  under  constant  jpressure  of  gases,  is  the 
specific  lieat  determined  when  the  gas  is  permitted  to  expand 
under  the  constant  pressure.     We  denote  it  by  Cp. 

The  specific  heat  under  constant  volume  is  the  specific  heat 
determined  when  the  vohime  of  the  gas  remains  fixed.  AVe 
denote  it  by  c^. 

The  former  always  exceeds  the  latter,  for  in  addition  to 
increasing  the  ijnolecular  motion  of  the  gas,  external  work  is 
done  against  the  pressure  of  the  air,  expanding  it  about 
0.0020275  of  its  volume  for  each  degree,  as  shown  in  equation 
(431). 

If  the  same  quantity  of  heat  will  raise  one  pound  of  water 
m  degrees,  and  one  pound  of  any  other  substance  n  degrees, 
both  under  constant  pressure,  then 

m 
Cp=  —. 

The  following  are  the  values  of  Cp,  c^,  and  Cp  -^  c^  for  a  few 
gases : 

Gas.  ep  Cv  Cp  -i-cv 

Air 0.238  0.169  1.408 

Oxygen 0.218  0.156  1.400 

Hydrogen 3.405  2.410  1.412 

Steam 0.480  0.370  1.297 


247.  The  dynamical  specific  heat  is  the  specific  heat  ex- 
pressed in  foot-pounds  of  work.  If  ^  be  the  dynamical 
specific  heat  under  constant  volume,  and  ITp  that  under  con- 
stant pressure,  then 

ir„  =  Jc,y    Xp  =  Jcp;  (438) 

where  J=  772  foot-pounds,  or  424  kilogramme-metres. 

248.  Adiaba'hc  curve. — Let  /^  be  the  heat  energy  neces- 
sary to  change  the  temperature  of  a  unit  of  weight  of  a  gas  t 
degrees,  the  pressure  beings,  and  density  S,     Then 

q=Ap-S.r).  (439) 

Differentiating,  j^  being  constant,  and  reducing  by  the  aid 
of  (436),  gives 


880  GASES.  [248.] 

'  ~  \drj„  -  dt   dd-       r   dS'  ^^*^> 
and  differentiating,  considering  S  as  constant, 

c^  =  (f)^f.^^P.^.  (441) 


Letting 


r  =  ^,  (442) 


y  being  considered  constant,  (440)  and  (441)  give 

in  which  r  is  involved  only  implicitly.     Eliminating  r  from 
(439)  by  means  of  (434),  and 

^=/(i>-^),  (444) 

the  total  differential  of  which  is 

in  which  substitute  -^  from  (443),  and  we  find 


—  djp  —JpydS 


df^  1 


.••i^-t^X^);  (445) 

where/  is  an  arbitrary  function,  that  part  of  which  depend* 
ing  upon  ^  is  implicitly  a  function  of  ^  and  d,  as  shown  by 


[248.] 


OASES. 


381 


(444),  and  the  other  part,  S  -^p~T  ^  is  explicitly  a  function  of 
the  same  quantities.  The  product  of  the  two  parts  is  some 
function  of  the  same  quantities  expressed  in  terms  of  the 
ratio  ^  -7-  S  ;  the  entire  transformation  being  made  so  that  an 
integral  may  be  obtained,  though  it  be  in  a  functional  form, 
i^^ represents  some  other  arbitrary  function,  and  q)  the  inverse 
of  it.  ^ 

In  the  particular  case  where  q  remams  constant^  as  it  would 
when  a  gas  is  compressed  or  dilated  in  a  close  vessel  imper- 
vious to  the  transmission  of  heat  through  its  walls,  although 
the  indicated  temperature  would  change,  we  would  have  for 
another  pressure  ^o>  ^^^  density  (Jo, 

and  eliminating  (p{q)  between  these  equations,  we  have 

jp6l=p,6y\  (446) 

and  since  the  volumes  will  be  inversely  as  the  densities,  we 
also  have 

pvPt  =  PqV^  =  constant^  (447) 

and  from  (436),  (446),  (447),  we  have 


.7.=ar=©'=(vr'=©'"  («») 


Fio.  183. 


Equation  (447)  is  the  sim- 
plest form  of  the  equation 
of  the  Adiabatic  curve,  or 
"curve  of  no  transmission" 
(from  the  Greek  ar,  not,  and 
Stapaiyetv,  to  pass  through), 
and  is  represented  by  the 
curve  AB,  If  an  isothermal 
line  pass  through  the  point 
A,  where  the  volume  of  the 
gas  is  1  and  pressure  J9i,  it 


882  GASES.  [249,] 

will  pass  above  the  adiabatic  for  values  of  v  >  1,  and  under 
it  for  values  of  t;  <  1. 

Examples. 

1.  If  a  given  volume  of  air  has  a  temperature  of  85°  F., 
what  will  be  its  temperature  when  dilated  to  double  the  vol- 
ume, performing  work,  but  receiving  no  heat  ? 

We  have,  from  equation  (448), 


/2|;\0.<03_ 


461.2+  85^ 
461.2  +  t  ' 

...  j5  =  -  50°  F. 

2.  If  a  prism  of  air  having  a  tension  of  \\  atmospheres  at  a 
temperature  of  88°  F.  expands  adiabatically  to  a  tension  of 
one  atmosphere,  required  the  final  temperature. 

We  have 

1  y^^     461.2  +  t 


JiJ 


549.2 
t  =  24°.35  F. 


3.  Required  the  ratio  of  the  volumes  before  and  after  ex- 
pansion in  the  preceding  example.  Ans.  1.335. 

4.  If  air  be  compressed  adiabatically  from  a  tension  of  15 
pounds  at  50°  F.  to  that  of  90  pounds,  required  the  final  .tem- 
perature. 

5.  If  air  at  60°  F.  and  six  atmospheres  expands  adiabati- 
cally to  one  atmosphere,  required  the  final  temperature. 

249.  Telocity  of  a  wave  ln^  an  elastic  medium."^    Assume 

*  The  general  problem  of  wave  propagation  lias  received  the  attention  of 
several  of  the  most  eminent  mathematicians  since  the  days  of  Newton,  and 
many  problems  have  been  solved  in  a  satisfactory  manner.  The  simple 
method  of  Newton,  Principia,  Prob's  XLIII.-L,  B.  II,,  has  not  been  excelled, 
and  the  definite  theoretical  result  obtained  is  quoted  to  the  present  day, 
although  the  effect  of  heat  upon  the  velocity  of  sound  was  not  then  known. 
La  Place,  in  the  Mecanique  Cileste,  tomes  II.  and  V.,  has  treated  of  the  oscil- 
lations of  the  sea  and  atmosphere  ;  Lagrange,  in  the  Mecanique  Analytique, 


[249.]  GASES.  883 

that  the  medium  is  confined  in  a  prismatic  tube  of  section 
unity,  E  the  coefiicient  of  elasticity  for  compression,  p  a  force 
which  will  produce  a  compression  dy  in  a  length  dx^  then 
from  definition  we  have 

The  lamina  dx  will  be  urged  forward — or  backward — by  the 
difference  of  the  elastic  forces  on  opposite  sides  of  it,  and  as 
the  quantities  are  infinitesimal,  this  difference  will  be  dj) ;  or 

Let  D  be  the  density  of  the  lamina,  then  its  mass  will  be 
M—  DdXy  and  we  have  from  equation  (21),  page  18, 

^"^df'-^d^' 
or, 

^--^^  (m^ 


which  is  a  partial  differential  equation  of  the  motion  of  any 

dxdt 


lamina.   JjetE-^rD  =  a^,  and  adding  a  ^^^^^  to  both  members, 


we  have 


dt      \dt  dxj      dx   \dt  dxj 


tome  II ,  has  discussed  the  problem  of  the  movement  of  a  heavy  liquid  in  a 
very  long  canal  ;  M.  Navier  published  a  M^moire  on  the  flow  of  elastic 
fluids  in  pi[)es,  in  the  Acndemie  des  Sciences,  tome  IX.  ;  and  M.  Poisson 
wrote  several  Memoirs  on  the  propagation  of  wave  movements  in  an  elastic 
medium,  and  the  theory  of  sound,  for  which  see  Journal  de  Vfkole  Poly- 
technique,  14th  chapter,  and  of  the  AcadSmie  of  Sciences,  tomes  IT  nnd  X. 
These  eminent  mathematicians  established  the  basis  of  all  the  analysis  for 
the  solution  of  the  problem.  More  recently  we  have  M.  Lame's  Lemons  sur 
VKlasiicifS  des  Corps  polides,  and  Lord  Rayleigh's  Treatise  on  Sound,  both 
of  which  are  works  of  great  merit. 


384  GASES.  [249.] 

Let 
then 

where  the  parenthesis  indicates  a  partial  differential  coefficient 
and 

and  equations  (449),  (451),  (452),  give 

(dV\_     (dV\ 
\dt)~  \dx  )' 

The  total  differential  of  V  =f{x,  t)  is 
by  substituting  (452), 


and  integrating, 


r=Fi.+at)=%^a%  (453) 


where  7^  is  any  arbitrary  function. 

Similarly,  subtracting  a  -^^  from  (449), 


F'=/(.-.0==|-«|.  (454) 


[249.J  GASES.  385 

Adding  and  subtracting  (453)  and  (454),  and  we  have  the 
respective  equations 


S=r.^(^^^^)-ry(--^^)- 


But 


y  =/(-^>  0 ; 


and  substituting  from  above,  gives 

dy  =  —  F{x  +  at)d{x  -^  at)  —  -^  f{x  —  at)d{x  —  at)  ; 
2ta  uGL 

integrating, 

y  =  tp  (ps  +  at)  -\-  (p(x  —  at\  (455) 

where  ^  and  q>  are  any  arbitrary  functions  whatever.  Their 
character  and  initial  values  must  be  determined  from  the  con- 
ditions of  the  problem.  The  equation  represents  a  wave  both 
from  and  towards  the  origin.  If  the  wave  be  from  the  origin 
only,  the  function  may  be  suppressed,  and  we  have 

y  =  ^  (aj  4-  at),  (456) 

and  differentiating. 


(I)  =  n-  +  «o, 


which  is  the  rate  of  dilation  (the  expansion  or  contraction 
of  a  prism  of  the  air),  and 


(^)  =  « •  f{^  +  «0; 


85 


886  GASES.  [24».] 

which  is  the  velocity  of  a  particle,  and  dividing  the  latter  by 
the  former, 

J=«,  (457) 

which  is  the  velocity  of  the  wave ;  hence, 

''  =  «=/|/5-  («8) 

The  elasticity  of  air  equals  its  tension ;  hence  Up  be  the 
pressure  per  square  foot,  w  the  weight  of  a  cubic  foot,  and  IT 
the  height  of  a  homogeneous  atmosphere,  equation  (424),  then 


n=i/^=VffH';  (459) 


hence  the  velocity  of  sound  should  equal  the  velocity  of  a  body 
falling  through  a  height  equal  to  one-half  the  height  of  a  uniform 
atmosphere. 

This  principle  is  applicable  also  to  the  vibration  of  elastic 
cords,  and  it  is  found  that 

The  velocity  of  vibration  of  an  elastic  cord  equals  the  velocity 
of  a  body  falling  freely  through  a  height  equal  to  half  the  length 
of  the  same  cord  whose  weight  would  equal  the  tension. 

Similarly,  in  water  neither  too  shallow  nor  too  deep, 

The  velocity  of  waves  on  the  sea  equals  the  velocity  of  a 
body  falling  freely  through  a  height  equal  to  half  the  depth  of 
the  sea. 

It  has  been  assumed  that  ^and  D  remain  constant  in  wave 
motion  ;  but  it  was  long  since  known  that  the  results  given 
by  equation  (458)  for  gases  did  not  agree  with  those  found  by 
experiment,  and  La  Place  showed  that  the  elasticity  was 
increased  by  the  action  of  the  wave  due  to  compression.  It 
is  necessary,  therefore,  to  consider  that  the  expression  is  cor- 
rect only  for  ultimate  values  ;  or 


^  =  j/^-  (*«o) 


dD 


[250,  251.]  GASEa  387 

Since  Eoz  wy,  dE=  '^—dw  =  ^^dD, 
w  w 


u 


\/^=  V^^-  (461) 


250.  To  FIND  THE  VALUE  OF  y^  WO  liavo  fpoiii  equatioD 
(461), 

r=p.«';  (462) 

by  means  of  which  it  may  be  found  when  the  velocity  of 
sound  in  a  gas  of  given  weight  and  tension  is  known.  In 
this  way  the  values  of  y  have  been  found  for  a  variety  of  sub- 
stances, a  few  of  which  are  given  in  Article  246.  It  is  con- 
sidered as  constant  for  any  given  gas,  and  nearly  constant  for 
all  the  more  perfect  gases. 

251.  Kemarks.  It  is  difficult  to  find  the  specific  heat  of  a 
gas  under  constant  volume  by  direct  experiment,  but  by 
means  of  equations  (462)  and  (442)  it  may  be  readily  com- 
puted when  the  specific  heat  at  constant  pressure  is  known. 
In  this  way  the  values  of  (?„,  given  in  Article  246,  were  deter- 
mined. 

If  the  temperature  of  air  in  its  quiescent  state  is  uniform, 
the  tension  will  vary  as  the  density,  equation  (416) ;  hence 
p  rr  w  will  be  constant,  and  the  velocity  of  sound  in  any  gas 
will  be  the  same  at  all  temperatures  or  densities,  bearing  in 
mind  that  the  factor  y  appears  on  account  of  the  condensation 
resulting  from  the  transmission  of  the  wave.  But  if  the  tem- 
perature varies  on  account  of  external  causes,  we  have,  equa- 
tions (435)  and  (461), 


-=\/ 


<k^  ^'''^ 


in  which  the  velocity  is  made  to  depend  upon  the  absolute 
temperature.    If  Tq  =  493.2,  jPq  -f-  Wq  =  26,221  feet,  equation 


888  GASES.  [252.] 

(424),  and  g  =  32J ;  the  velocity  in  dry  air  at  any  tempera- 
ture r,  will  be 

-  y  SZIgZEM7=  1,089.6  |/Tfeet.    (464) 

Examples. 

1.  Required  the  velocity  of  sound  in  air  at  the  temperature 
of  0°  F. 

2.  Required  the  velocity  of  sound  in  air  at  95°  F. 

3.  If  the  weight  of  a  cubic  foot  of  hydrogen  at  the  tension 
of  the  air,  14.7  lbs.  and  32°  F.  be  0.005592,  required  the 
velocity  of  sound  in  it  at  80°  F.,  y  being  1.4. 

4.  Required  the  velocity  of  sonnd  along  a  steel  bar,  the 
coefficient  of  elasticity,  p,  being  29,000,000  pounds  per  square 
inch,  and  w  =  486  pounds. 


-=/s 


°°°'°°  «6'"  "  '"  =  "■*'''  "»'  I«"  ■•«'°<'' 


The  velocity  of  sound  in  solids  depends  much  upon  their 
homogeneity.  Experiment  shows  that  its  velocity  is  from 
four  to  sixteen  times  as  great  as  in  air.  The  above  equations 
are  no  more  than  approximately  correct  for  solids. 

252.  Velocity  of  discharge  of  gases  through  orifices. 
If  the  density  and  temperature  remained  uniform  during 
flow,  the  law  of  discharge  would  be  the  same  as  for  liquids, 
and  the  same  formula  would  apply,  and  we  would  have  for 
the  velocity,  v^  —  2^A,  where  A  would  be  the  reduced  head, 
being  the  height  of  a  prism  of  the  gas  of  uniform  density 
equal  to  the  density  at  the  orifice.  If  the  temperature  be 
uniform,  the  density  will  vary  according  to  Mariotte's  law,  or 

P  ~  ^  ^'     ^^^  '^^  ^®  ^^^  velocity,  then  will  the  mass  passing  a 

transverse  section  of  area  unity  in  a  unit  of  time  be  du^  and 
the  difference  in  pressures  on  opposite  sides  of  a  section  will 


[2521  GASES.  389 

be  <fp,  which  will  eqn^l  the  momentum  of  the  elementary 

mass,  or 

dp——  dvdu ;  (465) 

in  which  the  quantities  have  contrary  signs,  since  the  greater 
the  velocity,  the  less  the  difference  of  the  pressures.  Divid- 
ing bji?. 


integrating, 


^^  —       —  ttit'tt , 
P  Po 

Pi       ^Po 


where  p  is  the  tension  of  the  gas  in  the  reservoir,  jf?i  the  ex- 
ternal pressure,  and  if  the  gas  flows  into  the  atmosphere  it 
will  be  14.7  lbs.,  u^  the  velocity  of  exit,  and  u  the  initial 
velocity  in  the  receiver.  If  the  receiver  be  large,  and  the 
oriiice  small,  p  may  be  considered  constant  and  u  zero,  in 
which  case  we  have 


"'Vl^^^'l;'  ^'''^ 


in  which,  if  p^=  0,  ?/i  =  oo ,  or  the  velocity  of  a  gas  into  a 
vacuum  would  be  infinite  according  to  this  hypothesis. 

If  the  temperature  and  density  both  vary,  the  effective 
head  will  vary.  Letting  z  be  positive  downwards,  we  have, 
from  equation  (417), 

dp 
w 


=1 


Following  the  method  of  Joule  and  Thomson,  and  consider- 
ing gravity  as  constant,  w  will  be  a  function  of  p  and  y ;  and 
substituting  and  reducing  by  means  of  equations  (448),  we 
have 


^9  J/),  -^^'o    p^      y-l    Wo    TqL        TiJ' 


(468) 


in  which  pi  is  the  tension  within  the  reservoir,  and  p2  that 
just  outside ;  then  if  t^  =  0,  the  flow  will  be  into  a  vacuum, 


890  GASES.  [253,  254.] 

and  according  to  (468)  the  velocity  will  be  a  maximum,  the 
value  of  which  will  be 


which  is  a/ times  the  velocity  of  sound ;  and  for  air 

becomes 

V  =  2,413  a/  —  feet  per  second  ; 

and  if  the  air  be  32°  F.,  then 

V  =  2,413  feet  per  second, 
or  about  2.2  times  the  velocity  of  sound  in  air. 

253.  The  volume  of  gas  flowing  out  per  second  will  be 

Q  =  sv,  (470) 

where  s  is  the  area  of  the  contracted  section. 

254.  The  weight  of  gas  flowing  out  per  second  will  be, 
equations  (448),  (468),  470), 


1 


'^.£i(iiy.  (471) 

..«=.y[*i.^(i-f;)](i;)^.(«.) 


which  is  a  maximum  for 

v-i 


which  for  air  becomes 

-'  =  0.8306,     ^  =  0.527,     ^  =  0.6345. 
n  Pi  ^1 


APPENDIX    I. 


SOLUTIONS    OF    PROBLEMS. 


PEOBLEMS. 

CONSTRAINED   MOTION. 

1.  A  particle  is  placed  at  the  extremity  of  the  vertical  minor 
axis  of  a  smooth  ellipse,  and  is  just  disturbed;  show  that  if  it 
quit  the  ellipse  at  the  end  of  the  latus  rectum  the  eccentricity 
must  satisfy  the  equation  e*  +  5e*  +  3e"^  =  5. 

Let  V  be  the  velocity  of  the  particle  at  the  end  of  the  latus 
rectum,  then 

«''  =  2#  =  2«r(i-£).  (1) 

Also  in  equation  (148),  p.  193,  we  have 

X=0,     Y=--mg, 
From  the  equation  to  the  ellipse, 

dx  1        , 

-y-  =     ,  for  X  =  as, 

ds       Vl  +  e' 
hence, 

V*  ^ga{l-  e')  (1  +  e«)«  TTf^,  =  ^«  (1  -  e%       (2) 
Equating  (1)  and  (2), 


2ya  ( Vi  -  e'  -  (1  -  e') )  =  ga  (1  -  e'), 

or,  

2Vl  -e'=(3+  e')  (1  -  e')  ; 

hence, 

e*  4-  5e*  +  3e'  =  5. 

2.  A  particle  slides  down  the  upper  surface  of  a  frictionless 

393 


394 


PROBLEMS. 


wedge,  the  wedge  being  free  to  slide  on  a  frictionless  horizontal 
plane ;  determine  the  motions. 


Let  W  =  the  weight  of  the  body, 
W'=  the  weight  of  the  wedge, 
h    =  AC,h  =  AB,  6  =  ACB,  iV^=  normal  action  between 

the  body  and  wedge, 
R  =  the  vertical  reaction  of  the  plane  AB  on  which  the 
wedge  slides,  a  —  DB  the  distance  moved  by  the 
wedge,  h  —  a  =  AD,  A  CD  =  a. 
Take  the  origin  of  coordinates  at  C,  x  positive  to  the  right, 
and  y  positive  vertically  down. 

The  forces  acting  on  the  body  are  N  and  W,  and  on  the  wedge 
N,  W,  and  E.  The  angle  6,,  equations  (143),  is  —  6,  and 
6*^  =  90°  +  ^  ;  and  the  equations  of  motion  become 


—  ~  =  TTcos  90°  +  JV^cos  (-  d)  =  iVcos  6, 


9    df 


0) 


-    ^  =  Fsin  90°  +  iV^sin  (-  6)  =  W-  JV^sin  8  ;  (2) 


—  ^  =  TT'  cos  90°  +  JV  cos  (180  -  6)  +  E  cos  270° 


iV  cos  6, 


(3) 


W'd'y'  _ 


'-  =  W'  sin  90°  +  iV^sin  (180  -  6)  +  E  sin  270  =  0  ;  (4) 

which  is  zero,  since  the  plane  is  fixed,  and  hence  there  will  be 
no  motion  in  that  direction. 


CONSTRAINED   MOTION.  395 

From  equations  (1)  and  (3)  we  have 


integrating, 

^  dt  -       ^    dt' 

and  again, 

Wx=-  W'x' ; 

F(5-«)=^'ap 

Wl     ' 

(5) 

and 

(6) 

Let  8  be  the  distance  from  C  down  the  plane. 

then 

y     =  8  008  0, 

differentiating, 


reducing 


X    =  8  ain  d  —  j-y; 

d^y=  cos  ^  d^s, 

d^x—  sin  6  d^8  —  J-  d^y, 

=  (sin  ^  —  ^cos  6jd'8; 


which,  in  (1)  and  (2),  gives 


d's  _  qN  cos  6  __  TT  —  JV^  sin  ^  . 

dP~  ^(  .'    .      ^         T~       >rcos^      ^*       ^^^ 
fFf  sin  6  —  J-C08  6) 


Solving, 


TFTT-sin^  , 

~  TT'  sin'  e  +  (W+  W)  cos' 8'  ^°' 


and  JVis  constant. 


396  PROBLEMS. 

Integrating  (1),  (2),  and  (3),  gives 


r~  =  Wv.^gNcoBd.t, 

(9) 

■  -^^  ^  Wv,  =  g(W-Nsme)t, 

(10) 

W ^j'  =  W'v:  =-gNoosd  -t; 

(11) 

and  integrating  again,  gives 

Wx  =\gN'cose  •  P, 

(12) 

Wy   =ig(W-  Nime)t\ 

(13) 

W'x'=  -igNcosO  ■  f. 

(14) 

Making  x  —  b  —  ain  (12),  x'  =  —  a  in  (14),  and  we  find  equa- 
tions (5)  and  (6)  as  before.  Also  make  y  =  h  in.  (13),  and 
find  iVas  before.     Then  equations  (8)  and  (13)  give 


t 


if     ^^,  .  fcos^  d  -f  w-^^;-,  sin^  d)        (14' 
1/  g  cos'  ^  \  W  +  W  J        ^ 


Integrating  (1)  and  (2)  in  reference  to  x  and  y,  and  we  have 
for  the  velocity  with  which  the  particle  reaches  the  foot  of  the 
plane, 

/— ,  r                            TTTT'sin'  6  ni    ,, ,, 

=:A/2flrA    1 7        (15) 

From  (11)  find 
,__  /^-T  r WW  mi  d -|i 

The  velocity  with  which  the  particle  reaches  the  horizontal 


CONSTRAINED  MOTION.  897 

plane  is  readily  found  by  the  principles  of  energy.     Thus  the 
work  done  by  gravity  equals  }V7u     Hence  we  have 

Wh  =  i^v^  +  i^v";  (17) 

which,  reduced  by  the  aid  of  (16),  gives  (15), 
If  If  =  GO,  we  have 

N=  TTsin  d,  (18) 


t    =a/-^^  (19) 

1/    g  cos^  6'  ^    ' 

v=V^,  (20) 

V'  =0;  (21) 

which  are  the  formulas  for  the  motion  of  a  particle  down  a 
smooth  ^a:e^  plane. 
For  the  inclination  of  the  path  of  the  particle,  we  have 

,  h-a  W      ,       ^ 

A  comparison  of  (15)  and  (19)  shows  that  the  particle  acquires 
a  less  velocity  when  the  wedge  is  free  than  when  it  is  fixed. 

Questions. — 1.  Show  that  whatever  be  the  relative  weights 
of  the  body  and  wedge,  the  common  centre  of  gravity  of  the 
two  masses  will  remain  in  the  same  vertical  line  during  the 
motion  down  the  wedge.     (See  Art.  149.) 

2.  Is  the  principle  of  the  conservation  of  energy,  Article  151, 
illustrated  by  this  example  ? 

3.  What  is  the  relation  between  W  and  W,  that  will  give  a 
minimum  velocity  for  the  body  TT?    (Eq.  (15).) 

4.  Given  TT,  equation  (15),  is  it  possible  to  assign  such  a 
value  to  W  or  to  d,  or  to  both,  so  that  vwill  be  \V^i  J\/2^A, 
or  -  \/2^  ? 


398 


PROBLEMS. 


5.  Is  there  an  algebraic  minimum  to  the  normal  pressure, 
equation  (8) — or  an  algebraic  maximum  ? 

3.  A  homogeneous  cylinder  rolls  down  the  upper  surface  of  a 
wedge  without  slipping,  the  wedge  being  free  to  move  on  a  fric- 
tionless  horizontal  plane  ;  required  the  motion. 

The  problem  may  be  generalized,  providing  the  successive 
elements  of  contact  of  the  body  and  plane  shall  be  in  the  succes- 
sive elements  of  a  cir- 
cular cylinder  circum- 
scribing the  body.  The 
centre  of  gravity  will 
move  in  a  vertical 
plane,  and  the  body 
may  be  a  cylinder,  a 
screw,  a  coil  of  wire,  a 
sphere,  an  ellipsoid  of 
revolution,  a  volume  of 
revolution  having  two  circles  of  contact  with  the  centre  of 
gravity  between,  and  other  forms.  Boiling  may  be  secured  by 
a  string  (or  flexible  band)  wound  around  the  body,  one  end 
being  secured  at  the  upper  end  of  the  wedge,  at  C,  and  the 
other  to  the  body.  Let  W  =  weight  of  the  body,  h  its  radius  of 
gyration,  TF'=  weight  of  the  wedge,  N=  normal  reaction  be- 
tween the  body  and  plane,  T  the  tangential  action  producing 
rotation,  DB  ^  z,  AB  =  b,  AD^l-z,  AC=h,  ACB=d; 
origin  at  C ;  x  positive  to  the  right,  and  y  positive  downward, 
r  the  radius  of  the  circle  of  contact. 

The  forces  acting  on  the  body  are  gravit}^  TT,  acting  vertically 
downward,  the  normal  reaction,  iV",  acting  directly  away  from 
the  plane,  and  friction,  or  tension  of  the  string  i^— whichever 
produces  the  rotation — acting  upward  along  the  plane. 

For  the  equations  of  the  motion  of  the  centre  of  the  body  we 
have  (167), 


Wd^ 
Jdf 

Wd^ 
9  df 


TTcos  90°  +  iV^cos  (-  6)  +  Tcos  (270°  --  d),      (1) 


Tf  sin  90°  +  iTsin  (-  8)  +  Tsin  (270°  -  0),      (2) 


tl 


CONSTRAINED  MOTION. 


899 


The  forces  acting  upon  the  wedge  are  its  weight,  W,  acting 
vertically  downward,  the  normal  reaction,  JV',  of  the  horizontal 
plane  acting  vertically  upward,  the  normal  reaction,  N,  between 
the  body  and  wedge  acting  downward,  and  the  pull,  T,  acting 
along  the  upper  surface  of  the  plane  downward.  Hence  we 
have 

J!L.yL=:W'  cos  90°  +  N'  cos  270°  +  N co%  (180°  -  d) 
g    df 


+  Tcos  (90°-  6), 


(3) 


The  origin  of  moments  being  taken  at  the  centre  of  the  body, 
we  have,  equation  (152),for  the  moment  of  rotation  of  the  body. 


MJc 


df 


Tr. 


(4) 


These  equations  may  be  reduced  to 

—  S=-^cos^-  TBmS 
9  di" 

—  ^=  W- lysine -Tcosd 
9  df 

—  ^=  _iV^cos^+  rsin  e 
9   dt' 

g        dv 


Integrating  twice,  assuming  iVand  T' constant,  as  they  are, 
9 


(5) 


W  "1*      w 
9    Jo       9 

T  =  -  —  «  =  i  (-  -^cos  6*  +  ^sin  6)  t\ 


X 

9 


9 


(6) 

(7) 

(8) 

(10) 


400  PROBLEMS. 

When  the  body  rolls  completely  down  the  wedge,  we  have 
rcpz=CB=:i  V^M^  =  a  (say)  (11) 

which  in  the  preceding  equation  gives 

which  reduces  the  preceding  equations  to 

W{h  -z)=  yJV  cos  e.f--  W^a_^ne^  ^^g^ 

Wh  =  y{W-:N'sm  6)  f  -  ^^'"^f^^^      (14) 


W  z  —  i^iV^cos  S  '  f ^ ,  (15) 


Subtracting  (15)  from  (13),  gives 

Wh 


W+W" 


(16) 


7  ^'^  /im 


From  (13)  and  (17), 
from  (14), 


g  00%  0  N      N 


(say) 


L  r^  J^(r~i\^sin^)' 


CONSTRAINED  MOTION. 


401 


|/^ 


A  sin  6 


W 
2  W'h'a 


Wk'  cos  ff 


r^g(B  +  A8in6)      ^,  ^.^,  ^ 


W 


W  -{-   IV 


7  +  r'  cos'  ^  +  ;fc» 


'-[ 


A*rYcos,'e-h  IPrY-iWk'  ar'gcose{B-^Asmd)  +  4  TT'^Pfl'n « 


r*}r(/y  +  A  sin  ^) 


^ 


r'W 


r 


The  entire  work  done  by  gravity  is  Wh,  and  this  equals  the 
energy  of  the  body  due  to  the  velocity  of  the  centre,  plus  the 
energy  due  to  the  rotation  of  the  body,  plus  the  energy  due  to 
the  motion  of  the  wedge  ;  or 

Wh  =  \Mv'  +  iMk'Qj"  +  \M'v'\ 

4.  A  particle  slides  cloion  a  frictionless  arc  of  a  fixed  vertical 
circle  ;  required  the  motion. 

The  forces  acting  on  the  particle 
will  be  gravity,  W,  and  the  normal 
action,  iV,  between  the  body  and  arc 
acting  normally  outward,  and  will 
be  the  difference  between  the  nor- 
mal component  of  the  weight  and  

centrifugal  force.     Using  polar  co- 
ordinates, origin  at  the  centre,  0  the  angle  between  r,  the  radius 
of  the  arc,  and  a  vertical  diameter,  and  resolving  normally  and 
tangentially,  we  have 


9      dV 


W 


m 


Wcosff-N. 


(1) 


&) 


402  •  PROBLEMS. 

Integrating  (1)  gives 

r |,  =  -  2^  cos  ^J^=  2^  (1  -  cos  Q),  (3) 

which  is  independent  of  the  mass.     In  (2)  this  gives 

J\^=(3  cos6'-2)F:  (4) 

At  the  point  where  the  body  leaves  the  arc  N  —^,  for  which 
condition  (4)  gives 

cos  ^  =  f  ;  (5) 

.-.  <9  =  48°11'  +. 
The  velocity  at' this  point  is,  equation  (3), 

The  subsequent  motion  will  be  that  of  a  projectile,  having  the 
initial  velocity  of  (6)  and  angle  of  depression  of  48°  11'  +. 
The  normal  pressure,  N,  varies  inversely  as  ^,  as  shown  by  (4). 
To  find  the  time  of  the  contact,  we  have  from  (3), 


cos  d 


The  superior  limit  makes  ^  =  —  oo,  which  shows  that  the 
particle  will  not  start  from  rest  at  the  highest  point.  It  will  be 
necessary  either  to  give  it  an  initial  velocity,  or  place  it  at  a 
finite  distance  from  the  highest  point. 

5.  A  hody  rolls  doimi  the  arc  of  a  fixed  circle  without  sliding, 
starting  at  a  point  indefinitely  near  the  highest  point  of  the  arc  ; 
required  the  motion. 


CONSTRAINED  MOTION. 


403 


Let  0  be  the  centre  of 

i^ 

the  arc  OB,  G  the  centre 

1 

* 

of  the  rolling  body,    W  =^ 
weight  of  the  body,   R  = 
OD,  r  =  DC,  the  radius  of 

& 

\ 

the  circle   of  contact,  k  — 

/ 

N 

the  principal  radius  of  gyra-          / 
tion  in  reference  to  a  hori-         / 

e-/     ■ 
/      ^ 

\ 

zontal  axis  through  G,  0  = 

o 

1 

BOD.      Resolving    tangen- 
tially  and  normall}^  we  have        \ 
for  the  motion  of  the  cen-         \ 

/ 

tre  G, 

^ — 

Q'^^"^ 

—  (72  +  r)  ^  =  IT  sin  6  +  iVsin  180°  +  2^  sin  270°, 
—  {R  4-  r)(^  =  Wco8  0  +  JV^cos  180°  +  T  cos  270°, 


or 


W{R  +  r)(^'=  Wg  cos  6  -  Ng, 
And  for  rotation  we  have  (152), 


(1) 
(2) 

(3) 


where  (p  is  the  angle  which  some  fixed  line  GF  of  the  body 
makes  with  the  vertical  ;  thus  cp  will  be  zero  when  6  is  zero. 
Since  arc  UD  =  arc  BD,  and  DCW  =  6,  DOE  =  ^  -  6>,  we  have 


r(9~  e)  =  Rd; 
R  -\-r 


9 


6, 


Solving  these  equations  gives 


T  = 


fc»+  r' 


W  sin  6. 


(4) 
(5) 


404  PROBLEMS. 

JVr=:  ["3  COS  ^  +  p^^2  (1  -  COS  6)  -  2~]  W.  (6) 

When  equation  (6)  becomes  zero  the   body  leaves  the  arc. 
Making  N=0,  we  find 

If  the  body  be  a  sphere  we  have  P  =  fr'^,  and 
cos  6  =  ^; 
.-.  ^  =  53°  58'. 

To  find  the  velocity  of  the  centre  of  the  sphere  when  it  leaves 
the  arc,  substitute  Tfrom  (5)  in  (1)  and  we  find 

{R  +  r)^  =  ^ffsme,  (8) 

and  integrating  and  reducing,  we  find 

(7?  +  r)  J  =  V-V^(l-oos6*)(7J  +  r),  (9) 

which  is  the  velocity  at  any  distance  6  from  the  vertical.    Making 
cos  6  =  If,  we  have  for  the  required  velocity, 


V^g{B  +  r).  (10) 

To  find  the  angular  velocity  of  the  sphere  at  the  point  where 
it  leaves  the  arc,  substitute  dO  found  from  (4)  in  (9)  and  make 
cos  ^  =  If  ;  we  thus  find 


V' 


-"i/'S.C*')-  "" 


From  this  point  the  sphere  will  continue  with  the  urfiform 
angular  velocity  given  in  (11),  (the  body  having  rolled  by  fric- 
tion, or  at  that  point  being  freed  from  the  string),  and  with  the 
initial  velocity  given  by  (10),  the  centre  moving  as  a  projectile. 
The  sphere  will  strike  the  horizontal  plane  tangent  at  the  lowest 


CONSTRAINED   MQTION.  405 

point  of  the  complete  circular  arc,  when  its  centre  is  at  a  dis- 
tance r  above  that  plane  ;  hence  the  sphere  will  be  a  projectile 
through  the  height 

%  =  i^cos  cos"^  ^  +  B  —r 

=  ^B-r.  (12) 

The  time  of  movement  as  a  projectile  will  be  given  by  equa- 
tions (a),  p.  177,  and  will  be  given  by  the  equation 

t'  +  0.2254  V'BlTr  -  t  =  0.10492i?  -  0.062177r. 
During  contact  it  will  rotate  on  its  axis 

R  ,10 

jr COS-1  —5 

2;rr  17 

times,  and  as  a  projectile  it  will  rotate 

oot  ^. 
^r—  times. 

If  the  arc  be  not  infinitely  rough,  let  /i  be  the  coefficient  of 
friction,  rotation- being  caused  entirely  by  friction  ;  required  the 
motion  for  this  condition. 

The  friction  will  be  //iV,  and  so  long  as  2^<  ^^iVthe  body  will 
roll,  but  from  the  point  where  T  =  juN  the  body  will  slide  on 
the  arc,  and  the  force  producing  rotation  will  be  ^^,  The  first 
part  of  the  motion  will  be  given  by  equations  (1),  (2),  and  (3) ; 
but  during  the  latter  part  /uN  must  be  substituted  for  T  in  (1) 
and  (3),  and  the  equations  integrated  again,  the  inferior  limits 
for  6  and  (p  being  the  values  found  from  (4),  (5),  (6),  and  T  = 
/iiN.  During  this  part  of  the  motion,  equation  (4)  will  not  be 
true,  but  instead  thereof,  if  5  =  rep'  be  the  total  arc  slip2)ed  over, 
we  have 

Re=r{cp-  d  ^  cp'),  (13) 

That  there  be  no  slipping  we  must  have 

or  from  equations  (5)  and  (6), 
ji-j-p  sin  e  <  //  ^3  cos  e  +  j^,  (1  -  con  6)  -2j.     (14) 


406  PROBLEMS. 

Equation  (3)  becomes 


f. -5  =  ....,  (15) 


and  (1), 

dt 


W{R  +  r)  -f-l  =  Wg  sin  0  -  ,xNg,  (16) 


and  (2)  remains, 

W{R  +  r)  (^y=  Wg  cos  6  -  JVg,  '(17) 

Eliminating  i\^  between  (16)  and  (17)  gives 

W{R  +  r)\^j^-fx  (^- j  J   =  TI  (/  (sin  ^  -  /i  cos  ^), 

which,  if  it  could  be  integrated,  would  give  d  =f{t),  and  then 
-TT  in  (17)  would  give  jV,  which  in  (15)  makes  known,  by  inte- 
gration, cp  =  F{t)  ;  and  these  results  combined  with  (13)  would 
make  known  cp'  in  terms  of  t.  The  time  of  the  movement  and 
the  superior  limit  of  6  during  contact  would  be  found  by  making 
J\^=  0  in  the  value  found  for  JSf.  If  the  body  had  a  rolling 
motion  only,  it  would  leave  the  arc  when  d  —  53°  58',  as  shown 
by  (7)  above,  and  if  it  slid  off  without  friction,  starting  from 
the  highest  point,  it  would  leave  it  at  ^  =  48°  11',  as  shown  by 
the  preceding  problem.  If  the  body  both  slips  and  rolls,  it  will 
leave  it  at  a  point  whose  angular  distance  from  the  vertical  is 
less  than  54°,  and  greater  than  48°. 

6.  What  must  be  the  radius  of  the  rolling  body  so  that  it  will 
touch  the  horizontal  plane  at  the  instant  it  leaves  the  arc,  after 
having  rolled  from  the  highest  point  ? 

7.  What  is  the  relation  between  the  radius  of  the  body  and  of 
the  arc  that  the  body  shall  roll  twice  on  the  arc  from  the  highest 
point,  before  leaving  it  ? 

8.  What  must  be  the  relation  between  the  radius  of  the  arc 
and  that  of  the  sphere,  so  that  by  rolling  from  the  highest  point 
of  the  arc  to  tlie  horizontal  plane  through  the  lowest  point  of 
the  circular  arc  the  body  will  rotate  once,  or  twice,  or  n  times  ? 


CONSTRAINED  MOTION. 


407 


9.  The  base  of  a  hemisphere  rests  on  a  horizontal  plane,  and 
a  sphere  rests  at  its  highest  point ;  if,  from  a  slight  disturbance, 
the  sphere  rolls  off  the  hemisphere,  how  far  from  the  centre  of 
the  hemisphere  will  it  strike  the  horizontal  plane,  and  what  will 
be  the  angular  distance  from  the  point  where  the  sphere  strikes 
the  plane  to  the  point  on  the  sphere  originally  in  contact  with 
the  hemisphere. — (Solution  in  The  Mathematical  Visitor y  Jan., 
1881,  p.  191.) 

10.  Oiie  sphere  rolls  down  another,  the  latter  leing  free  to  roll 
on  a  horizontal  plane  ;  required  the  motions. 

Let  R  be  the  radius  of  the  lower  sphere,  M  its  mass,  6  the 
angle  through  which  it  will  have  rolled  in  time  t  from  a  fixed 
vertical  line ;  r,  m,  d\ 
corresponding  values  for 
the  upper  sphere,  q)  the 
angle  between  the  line 
of  the  centres  and  the 
vertical,  T'  the  tangen- 
tial action  between  the 
spheres,  T  the  friction 
on  the  horizontal  plane, 
and  N  the  normal  action 
between  the  spheres. 

Take  the  origin  of  co- 
ordinates at  the  centre 
of  the  lower  sphere,  x 
horizontal  and  positive 
to  the  right,  y  vertically  upward,  and  /?  the  initial  angle  of  q). 
For  the  sake  of  distinction,  let  x'  and  y'  apply  to  the  upper 
sphere.  The  figure  represents  the  condition  of  the  bodies  at  the 
end  of  time  t. 

The  tangential  action  between  the  spheres  will,  at  first,  act 
downward  in  reference  to  the  lower  sphere  and  upward  in 
reference  to  the  upper  sphere.  The  normal  action,  JV,  will  not 
produce  rolling  of  the  upper  sphere,  but  will  tend  to  produce 
rolling  of  the  lower  one  in  an  opposite  direction  to  that  of  the 
tangential  action.  Any  modification  of  these  assumptions  due 
to  the  motion  will  appear  in  the  solution. 

For  the  lower  sphere  we  have 


408 


PROBLEMS. 


M^  =:JVcos  (270°  +  cp)A-T'  COS  (360°-^)  +  ^cos  180^ 
=  —  iV^sin  cp  +  T'  cos  g)  —  T 

and  for  the  upper  sphere, 

m -jj^  =  N%m.  g)—  T'  cos  cp 


;(i) 


m 


df 


N  cos  q)  -{■  T  mi  cp  —  mg 


Uir—  =  T' 


(3) 


and  for  the  geometrical  conditions, 

x  =  Re',  dx=z  Rdd,  (3) 

x'—  x  =  {R-]-r)  sin  q)',  .'.  dx'  —dx={R  +  r)  cos  cpdq),  (4) 

y'  =  (R-\- r) cos  cp]  .'.  dy'=—{R-\-r)  sin  cpdq?,     (5) 

R^cp^e-fi;)  ^  r{d'-(p  +  p) ;  R(dcp-dff)-^r{dd' -d(p).         (6) 

Eliminating  N,  T,  and  T'  from  equations  (1)  and  (2),  gives 


Integrating  once,  the  initial  values  being  0,  gives 


,,6Zar  do^      „      <?^' 


*^4' ' 


(7) 


,(8) 


from  which  we  see  that  whe?i  -^  =  -^rrrr  •  -rr  >  z^'^  ^«^^  MV  = 

dt       MR     dt  ' 

—  mv  J  or  the  moments  are  numerically  equal,  and  the  centres 


CONSTRAINED   MOTION.  409 

are  moving  in  opposite  directions,  ( V  and  v  being  respectively 
the  velocities  of  the  centres). 

Eliminating  dd,  dd',  and  dx'  by  means  of  (3),  (4),  and  (6), 
gives 

Ki/'+m)g  =  m(iJ  +  r)(f-co8^)J,  (9) 

and 

J(Jfrw)^  =Kf-cos  9>)+i(Jf+w)cos  (p]{R-^r)^.  (10) 

From  (9)  it  appears  that  the  motion  of  the  centre  of  the  lower 
sphere  will  he  positive  so  long  as  cos  (p  is  less  than  |,  or  (p  <  66° 
26'.  Should  the  spheres  separate  at  a  less  angle,  they  will  con- 
tinue to  roll  in  the  same  direction. 

From  (9)  it  appears  that  when  cos  6  =  \,  the  centre  of  the 
lower  sphere  will  be  at  rest,  in  which  case  (3)  shows  that  there 
will  be  no  rotary  motion ;  in  short,  the  lower  sphere  will  then 
be  at  rest.  Equation  (10)  shows  that  the  motion  of  the  upper 
sphere  will  be  continually  positive.  By  means  of  the  equations 
given  above,  a  complete  solution  may  be  found,  giving  the  nor- 
mal reactions  and  angular  velocities. 

11.  J  uniform  rod  of  length  a,  capable  of  making  complete 
revolutions  in  a  vertical  plane  about  one  extremity,  is  placed  in  a 
vertical  position  with  its  free  end  upiuard,  and  being  slightly 
displaced,  moves  from  rest ;  find  the  time  of  revolving  from  an 
angle  p  to  an  angle  6. 

The  equation  of  motion  is.  Art.  126, 

or 

^ad'O       .     . 
3  g  dV 

Multiplying  both  members  by  "Udd  and  integrating  between 
the  limits  6  and  0,  we  have 


410  PROBLEMS. 


or 


dt       l/2a\i  ,„ 

Hence  the  time  from  /?  to  ^  is 


= ©*■«' 


tanj^ 
tan  i>^' 


12.  A  rod  rests  with  one  extremity  on  a  smooth  plane  and  the 
other  against  a  smooth  vertical  wall  at  an  inclination  a  to  the 
horizon.  If  it  then  slips  down,  show  that  it  will  leave  the  wall 
when  its  inclination  is  sin  "\f  sin  a). 

Let  the  mass  of  the  rod  be  m,  its  length  2?,  its  inclination  to 
the  horizon  6,  and  the  coordinates  of  its  centre  of  gravity  x  and 
y ;  the  origin  being  such  that  for  the  time,  t,  considered, 
a;  =  ?  cos  ^  and  ?/  =  ?  sin  6.  Let  the  horizontal  and  vertical 
reactions  at  the  ends  of  the  rod  be  5"  and  F  respectively.  Then 
the  equations  of  motion  are 

d^y         ,c?^rsin  ff)  ^r  /-.n 

m^:=^ml     ^^^,    l^^my+V,  (1) 


d^x         ,  d^  (cos  6)       „  ._. 

\m^^  =  Hl  sin  6-  VI cos  d,  (3) 

Multiply  (1)  by  cos  <9,  (2)  by  -  sin  6,  and  (3)  by  (1  ^  Z);  and 
add  the  products.     There  results 

inil^  =  ^  my  COS  0,  (4) 


CONSTRAINED   MOTION.  * 


411 


whence,  since  ^r  =  0  and  6  =  a  when  ^  =  0, 


C5) 


The  rod  will  leave  the  vertical  wall  when 


=  0. 


Substituting  in  this  the  values  of  ;^  and  -j-  given  by  (4) 


and  (5)  we  have 


dt' 


^  =  sin-*(|  sin  «). 

{The  Analyst,  1882,  p.  193.) 


13.  An  angular  velocity  having  leen  impressed  on  a  hetero- 
geneous sphere,  about  an  axis  perpendicular  to  the  vertical  plane 
which  contains  its  centre  of  gravity  G,  and  geometrical  centre  C, 
and  passing  through  G,  it  is  then  placed  on  a  smooth  horizontal 
pla7ie.  Find  the  magnitude  of  the  impressed  angular  velocity 
that  G  may  rise  into  a  point  in  the  vertical  line  SCK  through 
C,  and  there  rest ;  the  angle  GCS  being  a  at  the  beginning  of  the 
motion,  a  the  radius,  and  q)  the  required  angular  velocity. 

Draw  the  radius  CGA,  and  from  O  drop  the  perpendicular 
GM  to  the  plane. 

Let  m  =  the  mass  of  the  sphere,  Ic  the 
radius  of  gyration  of  the  sphere  about 
an  axis  through  G  perpendicular  to  the 
plane  containing  C  and  G,  R  the  mutual 
reaction  of  the  sphere  and  the  plane, 
SM=  X,  GM  =y,  CS=  a,  angle  A  GM 
=  angle  ACS—  cp,  and  CG  —  c. 

Since  there  is  no  friction,  we  have  for  the  motion  of  the  centre 
of  the  sphere 


(1) 


41^  *  PROBLEMS. 

resolving  forces  yertically. 


m^  =  E-mg,  (2) 


and  taking  moments  about  G, 
d'cp  _ 


m¥  -^  —  —  Re  sin  (p,  (3) 

cp  being  the  angular  motion  of  the  sphere. 
We  have  y  =  a  —  c  cos  <p,  whence 

Substituting  in  (2), 

/     .        d^cp  dcp^        \ 

R  =1  mi^c  sm  cp-^  +  c  cos  ^-^  +  gj. 

This  in  (3)  gives  by  reduction, 

(g'  sin*^  cp  +  k')  -j^  +  c'  sin  (p  cos  cp -^- =  -  eg  sin  cp.    (4) 

Integrating, 

{c'  sin^  cp  +  ¥)^=C  +  2cg  cos  cp.  (5) 

Let  ^  =  0,  when  cp=:  a;  -~  =  go,  and 

C=  {c^  sin'  a  +  h'^co^  —  2cg  cos  a. 
Hence  (5)  becomes 

7       2 

(c'jsin''  cp  +  Jc^)  -~  =  (c*  sin*  « + ^)(i?'  +  2^:^  (cos  cp  —  cos  «').    (6) 


Now  if  the  initial  value  oi  cp  =  a,  the  terminal  value  —  Tt^a, 
bei 
and 


when  also  -^  =  0  ;  then  the  left  member  of  (6)  becomes  ^ero. 


»—     4p^  cos  a 
~~  c?^sin^ar  4-  Jc' 


{The  Analyst,  July,  1882.) 


KINETIC   ENERGY. 


418 


Kinetic  Energy. 

14:.  A  hall,  mass  m,  radius  r,  is  shot  with  a  velocity  v  into  a 
perfectly  hard,  smooth  tube  of  length  a,  radius  r',  mass  m',  free 
to  turn  about  its  middle  point,  which  is  fixed,  imparting  to  the 
tube  a  rotary  motion  ;  if  the  ball  just  reaches  the  centre  of  the 
tube,  required  the  angular  velocity  of  the  latter. 

In  this  case  the  kinetic  energy  of  the  ball  and  tube  due  to  the 
rotary  motion,  will  equal  the  kinetic  energy  of  the  ball  before  it 
enters  the  tube.  Let  k  and  Jc'  be  the  radii  of  gyration  of  the 
ball  and  tube  respectively,  and  gd  the  required  angular  velocity, 
then,  equation  (153), 

\mv^  =  imk'co^  +  i^'k''af ; 


0^ 


mv* 


m^  -f  m'k'^ 


If  the  tube  be  considered  slender,  we  will  have  k'*  =  -^a^ ;  also 
1^  =  |r*,  hence, 

60mv' 


c^  = 


24?»r'  +  5w'a' 


15.  A  cone,  mass  m  and  vertical  angle  2a,  is  perfectly  free  to 
move  about  its  axis,  and  has  a  fine,  perfectly  smooth  groove  cut  in 
its  surface,  making  constant  angle  fi  with  the  elements  of  the 
cone.  A  heavy  particle,  mass  M,  moves  along  the  groove  under 
the  action  of  gravity,  starting  at  a  distance  c  from  the  vertex; 
required  the  angle  through  which  the  C07ie  has  turned  when  the 
particle  is  at  a  distance  r  from  the  vertex. 

Let  p  be  any  variable  distance  from  the  ver- 
tex, (p  the  angle  through  which  the  particle 
has  moved,  k  the  radius  of  gyration  of  the 
cone,  and  6  the  required  angle. 

The  geometrical  relations  give 

p  sin  a      =  radius  of  the  cone  at  the  place  of 

the  particle  at  time  t, 
p  sin  adcp  =  the  horizontal  arc  through  which 

the  particle  moves  in  time  dt, 
dp  tan  /3  =  the  same  arc  ; 


414  PROBLEMS. 

.*.  dp  —  p  sin  a  cot  fidcp,  (1) 

The  moment  of  the  momentum  imparted  to  the  cone  in  an 
element  of  time,  will  be 

at 

The  angular  advance  of  the  particle  will  be  dq)  —  dd,  and  the 
horizontal  component  of  the  momentum  of  the  particle  will  be 

and  the  moment  of  the  momentum  will  be 

Since  gravity  has  no  horizontal  component,  the  horizontal 
motions  will  be  due  to  the  action  and  reaction  between  the 
bodies,  and  these  moments  must  be  equal ;  hence 

i.p»sin>.(^^>.4^  (.) 

Eliminating  dq)  by  means  of  (1)  and  (2),  we  have 

(    2 Jf  sin**  apdft  f c.    -  .   ^  in 

;,  .    .^  /.  o     =     2  sm  or  cot  fidQ ; 

fi  —  X  ^^^  ^  1      ^^^'^  +  ^r^  sin^  a 
~  ^ sin  a     ^ mF  +  Mc^  sin''  a' 

16.  An  elastic  ring,  mass  m,  natural  radius  a,  modulus  of 
elasticity  e,  is  stretched  around  a  cylinder ;  the  cylinder  sud- 
denly vanishes  ;  find  the  time  in  which  the  ring  will  collapse  to 
its  natural  length. 

'Tram 


r=|/^ 


17.  A  prismatic  har  in  a  vertical  position  rests  on  a  pivot  at 
its  lower  extremity  ;  it  is  slightly  disturbed,  required  its  kinetic 
energy  when  it  will  have  rotated  180°. 


KINETIC  ENERGY. 

We  have,  Prob.  19,  p.  216,  and  Art.  107, 


415 


dt' 


and  integrating 


ld(P     Sg        ^-]'     ^g 

2^  =  2r^^^n.=  'r' 


_!"' 


which  in  equation  (153)  gives 


Wl', 


that  is,  the  energy  is  the  same  as  if  the  bar  had  fallen  freely 
through  the  vertical  descent  of  the  centre  of  gravity  of  the  Mr. 

18.  If  a  sphere f  pivoted  on  a  horizontal  dia-    z 
meter  as  an  axis  without  friction,  oscillates  about 
an  external  axis ;   required   the  kinetic  energy 
when  vertically  under  the  support. 

Let  the  body  rotate  about  the  axis  of  y,  then 
will  equations  (177)  be  applicable,  and  we  shall 
have 

AB  =  1,  L,  =  0,  M,  =  -^Wl8m  <9,  uYi  =  0,Mg  =  W,x  =  Zsin  0, 
z  =  I  cos  6  ; 

.:  d'x=  —I  sin  6d0^  +  I  cos  dd'd, 
d'z^  -Icos  edS'-l  sin  ed'6 ; 

and  these,  in  equations  (177),  give 

d'd  9    •     n 

d^=-V'^''^ 

which,  integrated,  gives 


1  ^ 

2  dt 


^  =  |-co8^T=f-(l-cos^). 


The  velocity  of  the  centre  of  the  sphere  at  the  lowest  point 
will  be 


416  PKOBLEMS. 


,dd 


and  tlie  kinetic  energy  of  the  mass  when  it  rotates  through  180° 
will  be 

hence  in  this  case  also,  the  kinetic  energy  is  the  same  as  if  the 
sphere  had  fallen  freely  through  the  height  equal  to  the  descent 
of  the  centre  of  gravity. 

19.  Suppose,  in  the  preceding  example,  that  the  axis  of  the 
sphere  be  rigidly  connected  with  the  rod  AB ;  required  the 
velocity. 

Here  we  have,  as  in  example  19,  p.  216, 

d'd  gl       .     ^ 

1  dd"  gl      ,^  ., 

which,  compared  with  the  preceding  problem,  shows  that  tJie 
angular  velocity  will  be  less  at  the  lowest  point  when  the  sphere 
is  rigidly  con7iected  with  the  rod  A  B,  than  when  it  is  free  to  roll 
on  its  own  axis. 

The  kinetic  energy  in  this  case,  when  it  v/ill  have  rotated 
from  the  highest  to  the  lowest  point,  will  be,  equation  (153), 
page  202, 

i.m(r  +  |r^).|,  =  2TFZ; 

which  is  the  same  as  in  the  preceding  case.  The  time  of  vibra- 
tion will  be  greater  in  this  case  than  in  the  preceding — equation 
(161). 

Queries. — 1.  If  the  sphere  in  example  18  be  free  to  rotate  on  its  homon- 
tal  axis  as  a  diameter  while  the  entire  mass  rotates  about  an  external  axis, 
and  it  rotates  through  180^,  starting  with  no  velocity  from  a  point  vertically 
over  the  fixed  axis  ;  if,  at  the  lowest  point  in  its  path  its  axis  instantly 
becomes  rigid  with  the  bar  AB,  will  it  rise  to  the  highest  point  ? 


MOMENT  OF  THE  MOMENTUM.  417 

2.  In  the  preceding  example,  will  the  time  of  describing  the  second  part 
of  the  arc  be  the  Siime  as  that  of  describing  the  first  180°  ? 

3.  In  example  18,  if  tlie  sphere  gradually  melts  away,  will  the  velocity  or 
time  of  vibration  be  thereby  affected  ? 

4.  In  example  19,  if  the  sphere  gradually  melts  away,  will  the  time  or 
velocity  be  thereby  affected  ? 

5.  In  examples  18  and  19,  which  will  produce  the  greater  stress  on  the  axis 
of  suspension,  the  masses  and  arcs  of  vibration  being  the  same  in  both' 
cases  ? 

6.  At  what  points  of  the  rotating  masses  must  they  strike  a  fixed  obstacle 
so  a5  to  produce  no  shock  on  the  fixed  axis  ? 

7.  If  a  spherical  shell  B  be  rigidly  connected  to  the  bar  AB  and  filled  with 
a  frictionless  fluid,  would  the  time  of  vibration  and  the  velocity  at  the 
lowest  point  remain  the  same  if  the  fluid  should  suddenly  freeze  ? 

8.  If  a  spherical  shell  rigidly  connected  to  a  bar  and  filled  with  a  friction- 
less  fluid,  be  rotating  about  a  vertical  axis  with  a  uniform  velocity  under 
the  action  of  no  forces  ;  should  the  fluid  suddenly  freeze,  will  the  velocity 
of  rotation  remain  the  same  ?    Will  the  kinetic  energy  remain  the  same  ? 

Conservation  of  Areas  (Art.  150).     Moment  of  the  Mo- 
mentum (Art.  166). 

20.  A  cylinder  of  ice,  radius  r,  length  Z,  revolves  with  a  uni- 
form angular  velocity  gd  ;  if  it  he  subject  to  no  external  force 
and  melts,  required  the  angular  velocity  of  the  resulting  sphere. 

Neglecting  the  contraction  due  to  melting,  and  the  spheroidal 
form  due  to  rotation,  and  letting  oo'  =  the  required  angular 
velocity,  F  =  \E^  =  the  principal  radius  of  gyration,  we  have, 
Article  166, 

m  '  \r*Go  =  m  •  f -ff^cw', 

•  ,      5  r* 


4  R' 


To  find  E  we  have 


,      5  /4  r\f 


27 


41§  PROBLEMS. 

21.  If  n  spherical  shells  of  infinitesimal  thickness ,  mass  m 
of  each y  radius  r,  move  in  contact  without  friction,  having  their 
axes  of  rotation  all  in  one  plane,  the  angles  between  the  axes  and 

an  assumed  line  being  /5„  f3^,  A, A?  «^^  having  angular 

velocities  gdx,  00.^,  00^, g?„,  respectively,  suddenly  become  solid  ; 

required  the  position  of  the  resultant  axis,  and  the  resultant 
angular  velocity. 

Let  6  be  the  required  angle  and  oa  the  required  angular 
velocity ;  F  =  f  r^  the  principal  radius  of  gyration  before  ^nd 
after  becoming  solid  ;  then,  since  the  moment  of  the  momentum 
will  be  constant,  we  have,  resolving  parallel  and  perpendicular 
to  the  line  of  reference, 

nm¥ cos  6 •  Ga=m¥{oj^  cos /?i  +  ^^j cos /?2 4-  •  •  •  •  &?„  cos (3^=m¥A  ; 

nmy  sin  d  •  oDz=mk\Go^  sin  y^i  +  gl>,  sin  /^^ -f ....  &?„  sin  /?„)  =mk''B, 

where  A  and  B  are  the  values  respectively  of  the  parenthetical 
quantities.     Reducing,  we  have 


00  = 


VA'  +  B' 


n 


Q  1        ^  •      1         ^ 

d  =  COS"  ^ =r=rr-  =  SlU     ^  =  . 

VA'  +  B'  VA'  +  B' 

22.  A  prismatic  bar  rotating  in  free  space  suddenly  S7iaps 
asunder  at  its  centre  ;  required  the  subsequent  motion. 

The  body  will  rotate  about  its  centre,  and  after  separation 
each  half  will  rotate  about  its  own  centre,  and  those  centres  will 
have  the  same  uniform  velocity  that  they  had  immediately  pre- 
ceding separation.  After  separation  there  will  be  two  independ- 
ent systems,  still  at  the  instant  of  separation  the  entire  moment 
of  the  momentum  will  equal  that  of  the  original  bar. 

Let  /  be  the  length  of  bar,  m  its  mass,  ki  its  principal  radius 
of  gyration,  g?!  its  angular  velocity,  c«7  the  required  angular 
velocity  of  each  half  after  separation,  v  the  velocity  of  each  half, 
and  k  the  principal  radius  of  gyration  of  each  half;  then  we 
have 


also 


MOMENT  OF  THE  MOMENTUM.  419 

=  mh^cj  +  ^mlv  ; 

which,  substituted  in  the  preceding,  gives 

G?  =   Ce?i  ; 

hence  The  angular  velocity  of  each  half  after  separation  will  he 
the  same  as  that  of  the  original  bar,  and  the  two  halves  will  move 
in  opposite  directions  with  a  uniform  velocity.  Since  no  forces 
are  conceived  to  act  upon  the  bar,  the  kinetic  energy  after  sepa- 
ration will  be  the  same  as  before ;  hence  we  would  have,  equation 
(154), 

2(Jmi;')  +  \mh^GD''  =  ^mh\oo\ ; 
or 

(^looY  +  -hiW^'  =  ^noot ; 

as  before. 

Suppose  that  such  a  bar  separates  into  n  equal  parts,  what 
will  be  the  subsequent  motion  ?  Or  if  it  suddenly  melts,  or  dis- 
solves, will  the  elements  partake  of  the  rotary  motion  ?  If  two 
or  more  bodies  having  a  rotary  motion  cohere,  will  the  aggre- 
gate mass  have  a  rotary  motion  ?  Can  rotary  motion  impart  a 
motion  of  translation  ? 

23.  If  a  bar  rotating  about  one  end  suddenly  snaps  asunder, 
required  the  subsequent  motion. 

24.  If  a  bar  rotating  about  one  end  gradually  melts  away  at 
the  free  end,  will  the  angular  velocity  of  the  remaining  part  be 
changed  ? 

25.  A  spherical  shell  of  infinitesimal  thickness,  mass  m, 
radius  r,  is  filled  with  a  frictionless  fluid,  mass  m'  ;  the  shell  is 
rotating  with  an  angular  velocity  oo  when  the  fluid  becomes  solid 
and  rotates  with  the  shell ;  required  the  common  angular  velocity 
of  the  mass. 

Let  co'  =  the  required  angular  velocity.  The  moment  of  the 
momentum  of  the  shell  when  the  included  mass  is  a  fluid,  will 


420  PROBLEMS. 

be  f  mr^Go,  and  of  the  entire  mass  after  it  becomes  one  solid  will 
be  fmr'^G?'  +  m!¥oo' ; 


If  k'  =  lr\  then 


.'.   GO  =  -x — i m  (^' 


,  6m 

OOl  =  -z ji — ,  Q?. 

5m  +  3m 


G?  =  c». 


If  m'  =  0, 

as  it  should. 

26.  A  spherical  shell,  external  radius  r,  internal  radius  r^, 
mass  m,  filled  loith  a  fridionless  fluid,  mass  m',  rolls  on  a  i^er- 
fectly  rough  horizontal  plane  with  a  velocity  v  ;  the  fluid  freezes 
and  rolls  with  the  shell ;  required  the  velocity  v  of  the  common 
mass. 

Let  GO  be  the  angular  velocity  before  freezing,  and  gd'  after. 
The  moment  of  the  momentum  before  freezing  will  be 

mk^GD  +  (m  +  m')v  -r, 

and  this  will  equal  the  moment  of  the  momentum  after  freezing, 
hence 

m¥Go  +  (m  +  m')vr  =  mh^oo'  +  m'Tcloo'  +  (m  +  m')v'r, 

also 

2  /•'»  f^ 

V  =  rGD,    V  =  roj',      hi  =  ^r\,      ¥  =  -  ^,  _  ^j  ; 

/_  2m{r^  —  r?)  +  5(m  +  m'){r^  —  rfjr^ 

'  •  ^  ""  2tn  (r^  —  rl)  +  2m'rl  {r'  —  rX)  +  5  (m  +  m')  (r'  —  rl)r^ 

If  m'  =  0,  we  have  v'  =  v,  as  we  should.  The  kinetic  energy 
of  the  entire  mass  after  freezing  will  be  less  than  before ;  and, 
generally,  whenever  the  internal  forces  cause  a  relative  /change 
of  parts  or  particles  of  the  system,  the  kinetic  energy  may  be 
changed. 

37.  A  circle  is  revolving  freely  about  a  diameter  with  the 
angular  velocity  go,  when  a  point  in  the  extrernity  of  the  perpen- 


MOMENT  OF  THE  MOMENTUM.  421 

dicular  diameter  becomes  suddenly  fixed  j  required  the  instan- 
taneous angular  velocity  oo'. 

We  have 

mJc^i  GO  =  mk*Gj' ; 

28.  A  cone  revolves  about  its  axis  with  an  angular  velocity  go  ; 
the  altitude  contracts,  the  volume  remaining  constant,  required 
the  resultant  angular  velocity. 

Let  h  be  the  original  altitude,  x  any  subsequent  altitude,  r 
the  original  radius,  y  the  radius  when  the  altitude  is  x ;  then, 
the  volume  being  constant, 

\7tr*h  =  \7ty''x, 

the  moment  of  inertia  will  be 

mJc^  =  ^7cr*h, 

mh\  =  i^ny'x ; 

but  the  moment  of  the  momentum  being  constant, 

ml^x  ^  =  mT^GOx ; 

lc\  r"         x 

or  the  angular  velocity  will  vary  directly  as  the  altitude  of  the 
cone. 

29.  One  end  of  a  fine,  inextensible  string  is  attached  to  a  fixed 
point,  and  the  other  end  to  a  point  in  the  surface  of  a  homo- 
geneous sphere,  and  the  ends  brought  together,  the  centre  of  the 
sphere  being  in  a  horizontal  through  the  ends  of  the  string,  and 
the  slack  string  hanging  vertically.  The  sphere  is  let  fall  and 
an  angular  velocity  imparted  to  it  at  the  same  instant,  the  sphere 
winding  up  the  string  on  the  circumference  of  a  great  circle 
until  it  winds  up  all  the  slack,  when  it  suddenly  begins  to 
ascend,  winding  up  the  string,  the  sphere  returning  just  to  the 


422  PROBLEMS. 

starting  point  Required  the  initial  angular  velocity,  the  ten- 
sion of  the  string  during  the  ascent  of  the  sphere,  the  initial 
upward  velocity  of  the  centre  of  the  sphere,  and  the  time  of  move- 
ment. 

Let  I  =  the  length  of  the  string,  r  =  radius  of  the  sphere, 
m  =  its  mass,  ^  =  its  radius  of  gyration,  x  =  the  length  of  the 
unwound  portion  of  string  at  the  end  of  descent,  v  =  velocity  at 
the  end  of  descent,  v'  the  velocity  with  whicli  the  sphere  begins 
to  ascend,  00=  the  initial  angular  velocity,  gd'  =  the  angular 
velocity  with  which  the  sphere  begins  to  ascend,  t^  =  the  time 
of  descent,  t^  the  time  of  ascent,  T=  tension  of  the  string 
during  the  ascent,  and  /  the  impulse  communicated  by  the 
string  to  the  sphere  at  the  end  of  descent. 

The  length  of  string  wound  up  at  end  of  descent  is 


=''V" 


.  ,  2x 
root  I 

"     9 


tx  being  the  time  of  falling  freely  through  the  height  a?. 


/.  X  =  1  —  rcoi/- 


2x 

whence 

I  —  X  ,  /a 


(1) 


For  the  impulsive  motion, 

GO  —  CO    =— Ti, 

,       I 

V  +  V  =  — . 


Eliminating  /, 

But  ¥  =  \r\  and  v  =  ^/%gx, 


—  {go  —  go')  =  V  ■\'  v\ 

r  ^  ' 


:.  a,'  =  oo-^{V^  +  v').  (8) 


MOMENT  OF  THE  MOMENTUM.  423 

The  upward  motion,  the  origin  being  at  the  point  where  the 
centre  begins  to  ascend,  and  the  axis  of  y  positive  upwards, 

m^=T-mg.  (3) 

For  the  angular  acceleration, 

or 

Also 

y  =  rd, 
or 

d'y  _  rd^d 


df        dP 


(6) 


Eliminating  ^  and  ^  from  (3),  (4),  (5),  T=  fm^  =  |  the 

■tan 

weight  of  the  sphere.     Eliminating  T  and  -^  from  (3),  (4), 
(5),   -^  =  —  \g.      Integrating,   observing  that  when   /  =  0, 


df 
dt  ~^' 


%  =  v'-^t  (6) 


When  ^J  =  0,  ^  =  Jf, ; 

:.v'=^^  (7) 

Integrating  (6),  observing  that  when  ^  =  0,  y  =0, 

When  y  =  X,  t  =  Uy 

X  =  v%  -  -^gtl  (8) 


>4 

PROBLEMS. 

From  (7),  (8), 

v'  =  V'^gx, 

(9) 

'Vf- 

(10) 

From  (5), 

dy        dS 
dt  ~'^ dt' 

1 

v'  1=  tod',     oo'  —  -  V^-gx,  (11) 


Substituting  these  values  of  v'  and  go'  in  (2), 
Substituting  this  yalue  of  x  in  (1)  and  (9), 


eo  = 


_   (V^5-6)Vgi 


From  (10), 


^V[2(6-  V35)] 


'•yT7-=i/( 


14^(6  -  V35)\ 

^g        J' 


The  whole  time  is 


(Problem  by  the  Author  in  Mathematical  Visitor,  Jan.,  1879.) 

30.  If  n  concentric  uniform  spherical  shells  of  infinitesimal 
thickness  moving  loithout  friction  about  axes  whose  inclinations 
to  three  rectangular  axes  are  a^,  y5„  yx;  oc^,  y^g,  y-n  etc,  with 
angular  velocities  oo^,  ooi,  etc.,  respectively,  suddenly  become  one 


MOMENT  OF  THE  MOMENTUM.  426 

solid ;   required  the  resultant  angular  velocity  and  resultant 


axis. 


Let  a,  /?,  Yy  be  the  direction-angles,  gd  the  resultant  angular 
velocity,  h  the  radius  of  gyration,  and  m  the  mass  of  each  ; 
then.  Art.  1G6, 

nmk^oo  cos  a  =  mk'^[ojy  cos  a^  +  <»,  cos  or,  +,  etc.)  =  A,  (say) 

nm1(^Qi)  cos  p  =  mk^{cji  cos  /S^  -\-  gj^  cos  /?2  +,  etc.)  =  B, 

nmh^QO  cos  ;/  =  mk'^{ooy  cos  ;/i  +  gd^  cos  y,  +,  etc.)  =  (7; 

and  from  Coordinate  Geometry,  Art.  198,  Eq.  (3), 

cos'  a  +  cos'  p  +  cos*  ;/  =  1. 

Substituting  from  the  preceding,  we  have 


^  V^'  +  B*  4-  C'' 

This  value  in  each  of  the  preceding  values  gives 
cos  a  = 

cos  /?  = 


cos  y  =  — 7=-- 

If  the  motion  be  in  the  plane  xy,  we  will  have  y^  =  90°  =  ;/, 
=;/„  etc.;  hence  y  =  90°,  and  the  third  of  the  preceding  equa- 
tions vanishes,  and  the  case  reduces  to  the  problem  given  by  the 
author  in  the  Mathematical  Visitor,  Jan.,  1882,  p.  14. 

31.  A  screw  of  Archimedes  is  free  to  turn  about  its  axis  placed 
vertically  ;  a  particle  placed  at  the  upper  end  of  the  tube  runs 
down  through  it ;  determine  the  resultant  angular  velocity  im- 
parted to  the  tube. 


A 

\/A' 

+  B'+  C  ' 

B 

VA*  +  B'  +  (7'  ' 

C 

426  PEOBLEMS. 

* 

Let  m  =  the  mass  of  the  particle, 
nm  =  the  mass  of  the  screw, 

a  =  radius  of  the  screw, 

z  =  the  vertical  distance  the  particle  has  descended, 
and,  f^  and  cp  —  the  angles  through  which  the  screw  and  particle 
have  respectively  revolved  about  the  axis  in  the  time  t. 

The  energy  of  both  bodies  equals  that  imparted  by  gravity,  or 

ma'  -;^  +  ^  5^  +  ^^«  ^2  =  ^^9^-  (1) 

From  the  principle  of  Conservation  of  Areas,  we  have 

mar -—^  =mn  «  -rr  .  (*) 

dt  dt  ^  ' 

We  also  have  the  geometrical  equation 

z  =  a((p  +  6)  tan  a,  (3) 

From  (3), 

dz'        ,,     ,    fdcp'  ^^dcp   dd  ^dd'\  ... 

Substituting  in  (1), 

/     1_    dqP       sin'«   d(p    dd      sin''  a  dd'       cW\ 
""  y^^^'If'^'^coY^'  dt'dt'^  cos' a  dt'  "^""^f  y-'^^^-^^^ 

From  (2), 

dq)  _     dd 

~di~'^di' 


Substituting  in  (5), 


or. 


„f    n'  2n  sin'  oc      sin'  a        \  dd^ 

\cos'  a  cos*  a        cos''  a         /  <?^*         ^ 


a\n  -\-l)(n  +  sin'  a)  -^  =  2gz  cos'  a. 


MOMENT  OF  THE  MOMENTUM.  425^ 


When  —  =1  GDy  z  =  hf 


-y  {n  + 


2gh  cos*  a 
'.  oa  —  '  ' 


\){n  +  sin*  ay 


32.  A  cube  slides  down  an  inclined  plane  with  four  of  its 
edges  horizontal.  The  middle  point  of  its  lowest  edge  comes  in 
contact  with  a  small  fixed  obstacle  ;  determine  the  limiting  veloc- 
ity that  the  cube  may  be  on  the  point  of  overturning. 

Let  V  be  the  velocity  of  the 
cube  at  the  instant  of  impact, 
25  the  length  of  one  edge  of 
the  cube,  k  the  radius  of  gyra- 
tion in  reference  to  the  edge  at 
Pj  and  ^i  the  principal  radius 
of  gyration  in  reference  to  a  parallel  axis,  oo  the  initial  angular 
velocity,  and  m  the  mass  of  the  cube. 

The  moment  of  the  momentum  just  before  impact  will  be 

mv    b,  (1) 

t 
The  initial  moment  of  the  momentum  after  impact,  will  be, 
equations  (155)  and  (123),  and  example  4,  page  172, 

Qa  ==  mk'GD  =  m{W  +  k\)oj  =  m{W  +  |5')(» 

=  %m¥Go.  (2) 

Hence,  Article  166, 

mvb  =  |mJ*(i7 ; 

.-.  V  =  ^bo).  (3) 

The  cube  will  be  on  the  point  of  overturning  when  the  energy 
due  to  rotary  velocity  is  just  sufficient  to  raise  the  centre  to  a 
point  vertically  over  P.  The  energy  will  be,  equations  (153) 
and  (123), 

i^mr  .  GJ"  =  imk*  -  cy»  =  imoo^b'  =  ^mb'coK         (4) 
The  work  of  raising  the  centre  to  its  highest  point  will  be 


428 


PROBLEMS. 


wg  ■  Z'V'2[1  -  cos  (j.T  -  /3)], 
which  being  equal  to  (4),  we  have 

G7^  =  }V2  I  [1- cos  (iTT  -/?)], 

and  this  substituted  in  (3)  gives 

V'  =  -VV2  •  I?g[l  -  cos  (iTt  -  p)l 


(5) 


(6) 


(7) 


which  is  the  required  result. 

In  this  problem  we  may  consider  two  impulses,  one  that  of 
the  momentum  before  impact ;  the  other  that  destroyed  by  the 
impact.  It  is  now  required  to  find  the  magnitude  and  direction 
of  a  single  impulse,  which  applied  at  P,  will  produce  the  same 
effect.  At  first  it  seems  that  this  impulse  will  be  parallel  to  the 
plane  and  opposed  to  the  direction  of  motion,  but  if  the  body 
were  free,  it  would,  in  this  case,  rotate  about  its  centre,  equa- 
tions (168),  in  which  case  the  corner  at  F  would  move  perpen- 
dicularly to  the  diagonal  through  the  centre,  whereas,  in  the 
problem,  this  corner  becomes  instantaneously  fixed.  The  result- 
ant impulse  may  be  considered  as  the  resultant  of  an  impulse 
acting  along  the  plane,  and  another  acting  at  F  perpendicular 
to  the  diagonal  of  the  centre  through  that  point.  The  angular 
velocity  is,  equation  (3), 


CD 


3_v_ 

8i 


(8) 


hence  the  actual  velocity,  u,  of  the  centre  is 


'\/2b  •  00  =  u 


W-'iv 


which  is  also  the  initial  velocity 
with  which  the  point  F  would 
move  normally  to  the  diagonal  if 
the  body  were  free.  The  plane 
and  obstacle  then,  impose  the  two 

component  velocities  v  and  w,  the  angle  between  which  is  J;r ; 

hence  the  resultant  velocity  V  will  be 


MOMENT  OF  THE  MOMENTUM.  429 

F'  =  v'  +  w'  +  2vu  COS  |;r, 

.-.    V=W'^V'  (9) 

To  find  the  angle  between  V  and  v,  we  have 
w'  =  V^  -h  V-  —  2vV  cos  ^  ; 

.*.   cos  CD  =  —-=.,  (10) 

V34  ^    ' 

which  is  constant,  as  we  might  have  anticipated  from  (9),  since 
V  varies  directly  as  v.  To  find  the  energy  lost  by  the  impact, 
we  have,  for  the  kinetic  energy  before  impact,  equation  (24), 

imv' ;  (11) 

and  for  the  initial  energy  after  impact,  equation  (153),  problem 
4,  page  ]72,  equation  (123),  and  (8)  above 

=  i\rnv*,  (12) 

which  is  independent  of  the  dimensions  of  the  cube,  the  mass 
remaining  constant.  This  is  only  |  of  the  energy  in  the  cube 
before  impact,  hence  |  will  have  been  removed  and  passed  into 
heat. 

33.  A  circular  disc,  radius  r,  inass  m,  rolling  on  a  rough 
horizontal  plane  with  a  velocity  v,  impinges  against  an  obstacle 
whose  height  is  h  ;  if  there  be  no  slipping  on  the  obstacle,  required 
the  velocity  imynediatehj  after  impact,  the  height  to  which  the 
centre  of  the  disc  will  be  raised  if  it  does  not  pass  over  the 
obstacle,  and  the  velocity  at  the  highest  point  if  it  does  pass  over, 
and  the  velocity  of  approach  that  it  may  just  roll  over  the 
obstacle. 

The  velocity  immediately  after  impact  may  be  found  as  in 
Article  145,  and  the  other  results  as  in  the  preceding  problem. 


430 


PROBLEMS. 


34.  A  uniform  har  of  infinitesimal  section,  length  I,  mass  m, 
moving  with  a  uniform  velocity  v,  strikes  an  equal  har  at  rest 
lut  free  to  move  ;  if  the  two  bars  are  mutually  ^perpendicular , 
and  the  former  moves  in  a  path  perpendicular  to  both,  required 
the  motions  of  each  after  impact,  supposing  the  end  of  one  is 
struck  by  the  end  of  the  other,  both  bodies  being  inelastic.  What 
will  be  their  subsequent  motions  if  they  rigidly  adhere  at  the 

instant  of  impact  ? 

Let  OA  be  the  position  of  the 
moving  rod,  OB  that  of  the  rod 
at  rest,  at  the  instant  of  impact  ; 
OX  the  direction  in  which  0  was 
moving,  u  the  velocity  with  which 
the  centre,  C,  of  OA  was  moving 
before  impact,  v  its  velocity  after 
impact  in  the  direction  OX,  v'  the  velocity  with  which  the 
centre  D  of  the  other  rod  begins  to  move ;  h  the  radius  of 
gyration  of  each  rod  about  its  centre,  gj,  od',  their  respective 
angular  velocities  about  their  centres  ;  Q  the  impulsive  reaction 
at  0.     Put  il  =  a,  we  have 


mv  :-  mu  —  Q. 


m¥ 


Qa, 


mv'  =  Q. 
m¥oo'  =  Qa* 


(1) 

(2) 
(3) 
(4) 


The  velocity  at  0  of  the  moving  rod  after  impact  is  v  —  aai; 
that  of  the  other  rod  is  v'  +  aco'.  Since  the  rods  are  inelastic, 
these  velocities  are  equal,  hence 


v  —  acD  =  v'  +  aGo\ 
These  five  equations  readily  give 


(5) 


^=^'^*K^0- 


FRICTION.  431 

If  the  rods  rigidly  adhere  at  the  time  of  contact,  the  values  of 
r,  v'f  CO,  Gj',  found  above,  will  act  after  contact  on  the  connected 
rods.  Join  CD  ;  its  middle  point,  G,  will  be  the  centre  of  grav- 
ity of  the  connected  rods.  Join  OG.  Let  F=  the  velocity  of 
G  after  impact  in  the  direction  OX,  2m  V  =  mu  ;  .'.  V  =  \u. 
The  velocity  of  0  in  the  same  direction  is  v  —  aoa  =  \u.  Hence 
OG  will  move  in  the  direction  OX  with  a  velodity  =  \u, 

IT  1  ^' 


Hence  C  and  D  will  both  begin  to  revolve  about  G  with  a 

a* 
velocity  =  \u  — tj  f    The  angular  velocity  of  the  system  about 

^« = ^ ^ m- = Tf- H-^-    «"'^«"'"*'"^ ""  =  *"• ' 

3  u 

a  =  U\  v'  =  \Uy  00=  co'  =  -J- ,  and  the  angular  velocity  about 

OGy  when  the  rods  adhere,  will  be 

FRICTIOK. 

35.  Find  the  conditions  that  a  hoop  shall  roll  down  an  inclined 
plane  without  sliding,  //  being  the  coefficient  of  friction,  and  i 
the  inclination  of  the  plane. 

Let  6  be  the  angle  through  which  the  hoop  has  rolled  at  the 
time  t,  a  the  radius,  m  the  mass,  and  F  the  friction.  Then  we 
have.  Art.  125, 

a*m  -jjr  =  Fa, 
dt* 

and 

m   ^  ^  •  >  gm  sm  %  ^  F» 

therefore 

gm  sin  i  <  2F; 


432 

PROBLEMS. 

but 

Fzizgmpicos  i; 

therefore 

tan  i  <  2//. 

36.  A  sphere  having  a  rotary  motion  about  a  horizontal  axis 
is  placed  gently  on  a  rough  plane;  determine  its  motion. 

Let  W  =  the  weight  of  the  sphere,  /i 
the  coefficient  of  friction,  2^  the  tangen- 
tial action  due  to  friction. 

There  may  be  two  cases :  1st,  if  there 
be  slipping ;  and  2d,  if  there  be  no  slip- 
ping.  /In  the  former  case  T  =  juW. 

For  the  motion  of  the  centre  we  have 


T=mW, 


l  =  Md 


(1) 


and  for  the  rotary  motion. 


or 


}j.Wr, 


(2) 


Integrating,  we  have 


dx  .  , 

X  =  i^gt^  +  Cit  +  Ci 

1if<p=  —  i/jgrt'  +  c't  +  c" 


(3) 


For  ^  r=  0,  we  have  x  =  0,  and  cp  =  0  .*.  C2  =  0,  and  c"  =  0  ; 
dcp 


also  for  ^  =  0, 


dt 


(Wo,  the  initial  angular  velocity,  .*.  c'=^q7o; 


FKICTION. 


433 


dx 
and  —  =  0,  the  initial  velocity  of  the  centre,  and  the  corrected 


equations  become 

dx         ,                  ^ 

X  =  iMgt' 

dep               }igrt 

1  ugrl* 

(4) 


dx 
The  Telocity  of  the  point  of  contact  will  be  plus  -^ ,  and 

minus  r  -^ ;  hence 
dt 


dx         dm  ,  ixqrH 

di-^M=>'s^-'^^-]r' 


(5) 


and  so  long  as  this  is  finite  the  preceding  equations  will  hold 
true,  but  when  it  reduces  to  zero  the  conditions  change.  To 
find  when  they  change,  make  each  member  of  (5)  equal  to  zero, 
and  solve  for  t,  which  call  t^  ;  hence 


tr  = 


GDQr1<^ 


M^  +  n' 

The  left  member  of  (5)  becomes 


dx  dcp 
dt-''  dt' 

integrating, 
differentiating. 

x  =  r<p; 
d'x         d^cp  ^ 

(6) 

(7) 
(8) 
(9) 


which  are  the  equations  of  motion  when  there  is  no  slipping. 
The  value  of  T  is  not  e([ual  to  /i  TT  after  the  time  ^i,  and  to  find 
its  value  combine  the  first  of  (1)  and  (2)  with  (9),  giving 


434:  PROBLEMS. 

T=0;  (10) 


hence  it  requires  no  friction  to  cause  the  point  of  contact  to 
have  no  progressive  motion.  The  motion  is  the  same  as  if  the 
body  were  in  void  space  under  the  action  of  no  forces,  having  a 
uniform  motion  both  as  to  the  translation  of  the  centre,  and  of 
rotation  about  the  centre. 

The  total  amount  of  slipping  will  be 

rep  -  x  =  roDoh  -  iMSf  — -r-  t\  (H) 

37.  If  a  sphere,  radius  3  feet,  weight  20  pounds,  rotating  ten 
times  per  second,  be  placed  on  a  horizontal  plane  whose  coeffi- 
cient of  friction  is  ^^  ;  how  long  will  it  be  in  coming  to  a  uni- 
form velocity,  how  far  will  it  have  traveled,  how  much  will  it 
have  slipped,  what  will  be  the  uniform  velocity  of  the  centre, 
and  the  uniform  angular  velocity  ? 

38.  If  a  cylinder  have  the  same  amount  of  material,  diameter, 
and  rate  of  rotation,  as  the  sphere  in  the  preceding  example, 
and  placed  on  the  same  plane,  which  will  first  attain  a  uniform 
motion,  which  will  have  the  greater  uniform  velocity,  and  which 
will  have  slipped  most  ? 

39.  If  a  cylinder  whose  altitude  equals  the  diameter  of  the 
sphere  of  example  37,  the  same  amount  of  material  and  rate  of 
rotation,  be  placed  on  the  same  plane,  which  will  finally  attain 
the  greater  uniform  rotation  ? 

40.  What  must  be  the  coefficient  of  friction  that  there  be  no 
slipping  at  the  point  of  contact  at  the  beginning  of  motion  ? 

41.  Which  will  first  attain  a  uniform  motion,  the  sphere  in 
example  37,  or  a  sphere  of  the  same  material  and  twice  the 
diameter  ? 

42.  If  the  body  gradually  contracts,  retaining  a  constant  mass 
and  same  form,  will  it  go  farther  or  not  before  attaining  a  uni- 
form velocity  ? 

43.  If  a  sphere  preserves  the  same  circle  of  contact,  but  grad- 
ually contracts  laterally,  changing  to  an  oblate  ellipsoid,  will  it 


FRICTION.  435 

affect  the  time  of  attaining  a  uniform  velocity  ?    What  will  be 
the  time  if  the  polar  axis  becomes  half  the  equatorial  diameter  ? 

44.  A  rope  is  stretched  round  a  rough  cylindrical  surface  suit- 
tending  an  angle  0,  the  coefficient  of  friction  being  j^i ;  required 
the  force  F  acting  in  the  direction  of  the  tangent  at  one  end 
in  order  that  P  will  he  in  a  state  just  bordering  on  motion 
towards  F, 

Let  the  tension  at  any  point,  a,  be  t, 
that  adjacent,  h,  will  be  ^  +  dty  p  the 
normal  pressure  on  the  arc  per  unit  of 
length  if  it  were  uniform  ;  hence,  on  an 
element  of  length,  it  will  hopds,  fx  the 
coefficient  of  friction  ;  then 

dt  =  i-i 
also 


pds  =  \/i* 

+  (t  -{-  dty  +  2t(t  +  dt)  COS  {7t  -  6) 

=  t  V3(l  +  COS  (;r  -  6^)),  (ultimately) 

=  t'2 

COS  \(7t  —  6)  =  t  '^sin^e 

=  tdS  (ultimately)  ; 

;.  dt  =  ^At' 

dd. 

Integrating, 

But  for 
and  for 

log  jf  =  //^  4-  0. 
6  =  0,    t  =  F, 
e  =  AB,     t  =  P; 

.'.    P  =  i^£M«. 

Prom  the  relations 
and 
we  find 

pds  =  m, 

da  =  rdO, 

4 

t 

the  same  as  equation  {o),  p.  139. 


436  PROBLEMS. 

45.  A  shaft  having  a  bearing  the  entire  length  is  driven  by  a 
pulley  at  one  end,  the  power  being  taken  at  the  other.  Find 
the  diameter  of  the  shaft  at  any  point  for  uniform  strength. 

Let  w  =  the  weight  of  the  shaft  per  unit  volume,  pi  =  the 
coefficient  of  friction,  h  =  the  radius  at  the  driving  end,  x  the 
distance  of  any  cross  section  from  the  driving  end,  and  y  the 
radius  of  that  cross  section.     Then 

Wfxny'^dx  =■  friction  of  an  element, 

wpiTty^dx  —  moment  of  friction. 


Then 


wfxny^dx  +  PR  =  cy^,  (a) 


PR  being  the  moment  of  the  driving  power.     From  (a) 
wjXTiy^dx  =  Scy^dy, 


or,  letting 

W/ATT 


3c    -^' 


Jo  i,  y 


Ax  =  Xo^, 

y 

and 

y  =  A^-^^ 


ATTRACTION. 

46.  Assume  that  two  spheres  of  the  same  material  as  the  earth, 
each  one  foot  in  diameter,  are  reduced  in  size  to  a  mere  point  at 
their  centres,  and  placed  one  foot  from  each  other,  required  the 
time  it  would  take  for  them  to  come  together  hy  their  mutual 
attraction,  they  being  uninfluenced  hy  any  external  force. 

Let  B=:  the  mass  of  the  earth, 

m  —  the  mass  of  one  of  the  spheres, 
7n'  =  the  mass  of  the  other  sphere, 
R  =  the  radius  of  the  earth, 
r  =  the  radius  of  one  of  the  spheres. 


ATTRACTION.  437 

r'  =  the  radius  of  the  other  sphere, 
g  =  the  acceleration  due  to  gravity  on  the  earth, 
jA  =  the  acceleration  due  to  the  attraction  of  a  sphere  of 
mass  unity,  upon  another  sphere  of  mass  unity, 
the  distance  between  their  centres  being  unity, 
a  =  the  original  distance  between  the  centres  of  m  and  m', 

and 
X  =  the  distance  between  their  centres  at  the  end  of  time  t. 

Then  from  (227)  we  have 

which  integrated  (pp.  33,  34),  observing  that  for  ^  =  0,  a;  =  a, 
and  t;  =  0,  and  that  //  in  the  reference  equals  (pi  +  w')  — ~  in 
this  case,  gives 


../.).- 


i  =  To^^^^TW-T  X  \(ax  -x-)^^a  cos  M^- 
which  for  the  limits  gives 

,  _ .   r     ^(^     1^ 

If  both  spheres  are  of  the  same  density,  their  masses  will  be  as 
the  cubes  of  their  radii ;  or 


and  we  have 


.     ,     r      JRa      -\i 

and  if  the  spheres  are  equal,  as  in  the  problem,  we  have 
.      ,      /Ra\i 


488  PROBLEMS. 

Iia=^l  foot,  E  =  20,850,000,  r  =  J  foot,  g  =  32^,  we  have 
^        ^OK./I66,800,000\i 

=  1,788  seconds,  nearly, 
=  29.8  minutes,  nearly. 

47.  To  find  the  stress  in  pounds  which  would  be  exerted  by  the 
mutual  action  of  two  such  spheres  as  in  example  46,  at  a  dis- 
tance of  one  foot  between  their  coitres,  tve  have,  from  equations 
(224)  and  (235),  since  m  =  rn ,  and  x=l, 

64  M'  ' 

The  mass  of  the  earth  is  5|  times  an  equal  mass  of  water. 
The  weight  of  a  cubic  foot  of  water  is  62 J  lbs.  at  the  place  where 
g  =  32 J-,  and  the  yolume  of  the  earth  is  ^TrB^  hence 

£J=6i  X  62i  X  ^7tB\ 

which,  substituted  above,  gives  for  the  stress 

f TT  X  5|  X  62|  X  32j 
64  X  20,850,000 

=  .00003471+ lb. 

347 
or  nearly  of  a  pound,  a  quantity  inappreciably  small. 

48.  If  the  density  of  the  earth  at  the  surface  be  unity,  and  at 
the  centre  is  m,  g  the  force  of  gravity  at  the  surface,  and  f  that 
at  any  distance  x  from  the  centre  loithin  the  earth,  what  is  the 
law  for  f,  the  density  increasing  uniformly  toward  the  centre, 

•   The  density  at  any  distance,  x,  from  the  centre  will  be 

(m  —  1)  ,         .    ,  ^ 
^^^ ^  (r  —x)  +  1. 

r        ^  ' 


ATTRACTION.  439 


The  volume  of  a  sphere  of  radius  a;  is  4;r    xHx.     Hence  W( 


have, 


and 


Mass  =  ^ttI"  n^\^  ^^  (r-x)  +  llx^x, 
4;r  r(m  -  1)  [tt?       a;^        aT] 


=  4[^( 


m  ^-r  —  \)x  —  3(//i  —  1)^:*  |, 


If 

hence. 


x=r,    f  =  g, 

g  =  c—\  4(m  +  r  —  l)r  —  ^{m  —  l)r* 

_  4(m  4-  r  -  l)rg  -  3(m  -  l)x* 
•'•  •^~4(,m  +  r  -  l)r  -  3(m  -  l)r"*^' 

49.  Considering  the  earth  and  moon  as  uniform  bodies,  the 
mass  of  the  earth  5  J  times  that  of  an  equal  volume  of  ivater,  and 
its  radius  3,956  miles,  the  mass  of  the  moon  3  J  times  that  of  an 
equal  volume  of  water,  and  its  radius  1,080  miles,  the  mean  dis- 
tance between  the  centres  of  the  earth  and  moon  CO  tiines  the 
radius  of  the  earth,  and  the  acceleration  due  to  gravity  at  the 
surface  of  the  earth  32^  feet  j^er  seco?id ;  required  the  time  in 
which  they  tvould  come  together  by  their  mutual  attraction. 

When  they  are  in  contact  the  distance  between  their  centres 
will  be  5,036  miles,  and  we  have  from  Problem  46, 

'-t-prf^]'[<"--)*— ©']:,<■> 


But 


7i/"      R'  7  Mr' 


and 
t 


=  [.!^lVrJ['«-^'— (:-)*]!■« 


440.  PROBLEMS. 

Substituting  numerical  values  in  equation  (2), 


-".— (iS)'] 

—  412,945  seconds. 

Loomis  gives  415,600  seconds  as  the  time  required  for  a  par- 
ticle to  fall  from  the  moon  to  the  earth,  the  distance  to  the 
moon  being  the  same  as  that  given  in  this  example,  which  is 
2,655  seconds,  or  nearly  44  minutes  more  than  the  time  for 
the  earth  and  the  mooh  to  meet. 

50.  When  the  earth  is  in  perihelion,  suppose  the  sun's  mass  to 
he  increased  by  x  times  its  present  value.  Required  the  change 
in  the  elements  of  the  terrestrial  orhit. 

The  square  of  the  eccentricity  of  any  planetary  orbit  is  (Prob- 
lem 4,  page  189), 

.        „  Vr  sin*  /?        FV«  sin*  ft 
1  —  <*  '  4- 


in  which  V  is  the  velocity  of  the  planet  at  the  point  whose 
radius  vector  is  r,  /?  the  angle  between  the  curve  at  that  point 
and  r,  and  fj.  a  measure  of  the  attractive  force.  In  this  case 
a  =  90°,  and  the  equation  becomes,  by  reduction, 

FV      ,   ^ 
=1  ±e.     , 

If  now  the  mass  of  the  sun  is  increased  suddenly  x  times  its 
present  value,  /x  becomes  xja,  and  the  eccentricity  of  the  new 
orbit  will  be  ei  (say),  hence, 

V- 
and 


xju            -^    i> 

FV 

=^(i±.). 

-1, 

ATTRACTION.  441 

in  which  ei  must  be  used  when  the  right  member  of  the  equa- 
tion is  positive,  and  —  e^  when  it  is  negative. 

It  X  <  ^{1  ±  e),  Ci  >  1,  or  the  orbit  will  be  a  hyperbola. 
''  x  =  1(1  ±  6),  e,  =  1,         "         "         "        parabola. 

"  ^  I  ^^n±c)  I  '  ^>  <  1'  "         ''         "  a^  e^lips^- 
''  x=l  ±e,   ±e,  =  0  ''        "        "a  circle. 

*'  a;  >  1  ±  e,   —  e,  >  —  1,     ''         "         "   an  ellipse. 

In  all  but  the  last  the  given  and  resulting  perihelia  coincide ; 
but  in  the  last  the  given  perihelion  will  coincide  with  the  new 
aphelion,  x  =  ^  and  e  =  0.01G784,  the  eccentricity  of  the  earth's 
orbit,  we  find  —  Bi  =  —  0.32214,  or  —  ^i  =  0.34452  ;  hence  the 
present  perihelion  would  become  the  aphelion  point  of  the  new 
orbit,  and  the  new  orbit  would  be  an  ellipse  with  the  eccentric- 
ity 0.32214  or  0.34452.     The  mean  distance  would  become 

of  its  present  distance. 
The  time  of  rotation  about  the  sun  would  be 


X  365  J  days  =  191.25  days. 


61.  Suppose  infinitesimal  aerolites  equally  distributed  through 
all  space,  every tvhere  moving  equally  in  all  directions  with  a  uni- 
form  a7id  constant  absolute  velocity.  The  aggregate  mass  inter- 
cepted in  a  given  time  by  a  given  stationary  sphere  is  supposed 
to  be  known.  Determine  the  effect  upo7i  the  eccentricity  of  a 
spherical  planet  of  given  mass  and  volume  moving  in  an  eccentric 
orbit  all  of  whose  elements  are  known.  {Math.  Visitor,  July, 
1880.) 

If  the  velocity  of  the  planet  be  less  than  that  of  the  aerolites, 
the  same  mass  will  be  intercepted  as  if  the  planet  was  at  rest. 
Consider  this  case. 

The  change  of  the  elements  of  the  orbit  will  be  due  to  two 
causes.     1st.  The  increase  of  the  mass  of  the  planet  will  increase 


442  PEOBLEMS. 

the  attractive  force  between  the  sun  and  the  planet.  2d.  The 
aerolites  will  cause  a  direct  resistance  to  the  motion  of  the 
planet. 

Let  M  be  the  mass  of  the  sun,  m  the  initial  mass  of  the 
planet,  8  the  distance  between  them,  Tc  a  constant ;  then  will 
the  acceleration  of  one  body  in  reference  to  the  other  at  the  end 
of  time  t  (the  time  in  the  problem  being  unity)  be 

tcjU  -^  m'  +  mt) 

hence  at  distance  unity, 

fx  =  h{M  +  m'  +  mt)y 
and 

dpi  =.  Jcmdt, 

For  an  elliptical  orbit  we  have,  page  190, 

Vlrl  =  Ma{l  -  e')  =  c, 

a  being  the  semi-transverse  axis.  Since  the  changes  are  small, 
consider  two  quantities  only  to  vary  contemporaneously.  If  e 
be  constant,  we  find 

Tcmc 


da^ 

IV  nil/ 

^\1  -  e^) 

dt; 

Aa  = 

Tcmct 

nearly. 

Similarly,  if  a 

be  constant,  we  find 

therefore. 

de  = 

hn(l  -  e') 

2e/x 

dt; 

J,                   hmct  , 

ne  =  r nearly, 

'^H'  yajA,  {a/j.  —  c) 

Hence,  the  major  axis  decreases  and  the  eccentricity  increases 
with  the  time,  and  the  amount  of  change  for  one  revolution 
may  be  found  by  making  t  equal  to  the  corresponding  time. 

2d.  The  law  of  resistance  is  not  given.     The  aerolites  being 


ATTRACTION.  443 

infinitesimal,  we  do  not  consider  the  impact  as  between  finite 
masses,  but  they  constitute  a  medium  through  which  the  planet 
moves.  Considering  the  medium  as  of  uniform  density,  /),  and 
the  resistance  varying  with  the  square  of  the  velocity,  the  case 
comes  within  one  discussed  by  La  Place,  Mecanique  Celeste, 
(8925,  8926). 

The  equations  will  become  for  this  case, 


de  KDa  ,^ 

—  = au  ; 

e  fx 

K  being  a  consrant  depending  upon  the  mass  and  form  of  the 
planet.     If 

Aa 


1  +  "iK'a^e' 


Oi  and  Bi  being  initial.  The  major  axis  and  eccentricity  both 
decrease  as  the  vectorial  angle  increases,  and  the  orbit  becomes 
more  nearly  circular. 

The  plane  of  the  orbit  will  not  be  changed  ;  and,  finally,  the 
longitude  of  the  perihelion  will  not  be  changed. — Mecanique 
Celeste  (8916). 

52.  Show  that  if  two  bodies  revolve  about  a  centre,  acted  upon 
by  a  force  proportional  to  the  distance  from  the  centre,  and  iur 
dependent  of  the  mass  of  the  attracted  body,  each  will  appear  to 
the  other  to  inove  in  a  plane,  whatever  may  be  their  mutual 
attraction. 

Take  the  plane  xy  in  the  plane  of  the  central  force  and  of  the 
two  bodies  at  any  instant,  the  origin  at  the  central  force  and  the 
axis  of  X  passing  through  the  body  a,  the  coordinates  of  a  being 
x'  and  0,  and  of  the  other  body  b,  x"  and  y",  d  their  distance 


444  PROBLEMS. 

apart ;  Mthe  central  force  at  unit's  distance,  i^^the  force  of  b  on 
a,  and  F'  that  of  a  on  b  (according  to  the  Newtonian  law 
F=  F')  ;  X\  X",  Y',  Y",  the  axial  accelerations  of  a  and  b. 
Then 


X' 

-       Mx'  +  f'^"'''' 

—         irix    -h  -r           ,        , 

Y' 

-  F^"  ■ 

X" 

=      Mx"      J"^""'^', 

Y" 

=  -Mf-F'y^; 

Y"  -  Y'          y" 

which  gives  the  direction  of  the  relative  accelerations,  and 
which  is  parallel  to  the  line  ab.  Hence,  whatever  be  the  direc- 
tions of  motion  of  the  two  bodies  or  their  absolute  velocities, 
their  relative  positions  {a,  b),  their  relative  velocities,  and  their 
relative  accelerations  are  parallel  to  a  plane. 

Solution  by  Quaternions. — Let  p  and  pi  be  the  vectors  of  the 
two  bodies  referred  to  the  centre  as  origin.      Let  p'  =  -^ , 

/>i'  =  -^  ,  p"  =  -~-  ,  p^"  =  -jj- .     If  31  is  the  central  force  at 

the  unit  of  distance,  and  iV  and  JVi  the  mutual  attractions 
divided  by  the  distance  apart,  we  have 

p"  -  -  Mp  ■\-  N{p^  —  p), 
a"  =  -  i/pi  +  iViCp  -  a)  ; 
.-.  A"  -/>"=-  M{p,  -  p)  +  {N,  +  2i){p  -  p').  ' 
The  scalar  part  gives 

S[(p--pW-prW'-pr')>{^+^-^^r)S(p-p^y(p'-p\)=0, 


PROBLEMS.  445 


which  proves  that  the  apparent  orbit  is  a  plane. — (Coordinate 
Geometry,  p.  278,  eq.  (3).)     (Math,  Visitor ,  July,  1880.) 

53.  An  elastic  string,  without  weight  and  of  given  length,  has 
one  end  fixed  in  a  perfectly  smooth  horizontal  plane,  and  the 
other  to  a  point  in  the  surface  of  a  sphere,  the  string  leing  un- 
wound. The  sphere  is  projected  on  the  plane  from  the  fixed  point 
with  a  linear  velocity  v,  and  an  angular  velocity  od,  ivinding  the 
string  on  the  circumference  of  a  great  circle  ;  required  the  elon- 
gation of  the  string  when  fully  stretched,  and  the  subsequent 
motion  of  the  sphere. 

Let  r  =  the  radius  of  the  sphere,  a  =  the  original  length  of 
the  string,  oo  =  the  initial  angular  velocity  of  the  body,  i'  =  the 
initial  velocity  of  the  centre  of  the  body,  and  ti  =  the  time  of 
winding  the  slack.     Then 


\  +  r 

■cot, 

=  a 

> 

••A 

a 

• 

V 

+  roo' 

and  the  initial  stretched  part  will  be 

vfi  = =  I  (say). 

Immediately  following  this  time  the  string  will  be  stretched, 
and  the  tension  at 'first  diminishes  both  the  linear  and  angular 
velocities.  Take  the  origin  at  the  remote  end  of  I  for  the 
variable  motion.  Let  m  =  the  mass  of  the  body,  s  =  the  space 
passed  over  by  the  centre  during  time  t,  8  =  the  angular  dis- 
tance passed  by  the  initial  radius  in  the  same  time,  k  =  the 
radius  of  gyration  of  the  body,  e  =  the  coefficient  of  elasticity 
of  the  string,  ^  the  cross  section  of  the  string,  and  A  th.e  elonga- 
tion produced  by  the  tension  T  of  the  string.  Then  Mariotte's 
law  gives 

Assume  that  /  is  so  long  compared  to  rd,  that  the  latter  can 
be  neglected,  and  let  B  =  eA  -^  I,  then 

T  =  B\. 


446  PROBLEMS. 

The  conditions  of  the  problem  giye 

dX  =  ds-\-  rdO ;  (2) 

.-.  d'X  =  d's  +  rd'O.  '     . 

Also,  for  motion  of  the  centre, 

and  for  the  rotary  motion, 

mh'^^  =  -  Tr  =  -  BrX,  (4) 

which  two  equations  in  the  preceding  give 

Integrating,  observing  that  for  X  =0,  t  =  0,  and  dX  -^  dt 
zn  V  +  roOf  we  have 

A  =  — - —  sm  Dt.  (5) 

The  elongation  X  will  be  a  maximum  for  sin  Dt  =  1,  or 
t  =  7t  -7-  2D,  for  which 

The  time  of  producing  the  maximum  stretch  of  the  string  is 
independent  of  the  initial  motions.  When  the  string  returns  to 
its  original  length  X  will  again   be   zero,  and  sin  Dt  =  0,  or 

Dt  =  TT  ;    :.  t  —  j^. 

All  the  circumstances  of  the  variable  motion  may  be  deter- 
mined'  by  integrating  equations  (3)  and  (4).     Integrating  .after 

ds 
substituting  from  equation  (5),  observing  that  for  ^  =  0,  -^  =  v, 

5  =  0,  ^—  00^  and  ^  =  0,  we  have,  if  we  put  ^for  eA{y  4-  rco) 
~  mlD\ 


PROBLEMS.  447 

da 

^  =  Fico^Dt-l-\  +  v,  (6) 

s^^i&mDt-Dt']  +vt,  (7) 

dO  r 

^  =  ^p[cosi)<-l]  +  a.,  (8) 

Fr 

^  =  j)^ [sin  Dt  -  Dt]  +  cot.  (9) 

For  the  maximum  of  (5)  d\  ~-  dt  =  0,  which  in  (2)  gives 

d^_  __     d^ 
dt  ~      '^ dt' 

which   combined   with    (6)    and  (8)   gives   cos  /)^  —  1  =  ^  1 ; 

.*.   Dt  =  ^7t  as  before    found,  and  serves  as  a  check  upon  the 

work.      The  relation  ds  =  —  rdd  shows  that  the  direction  of 

one  of  the  motions  changes  sign.     At  the  point  where  the  linear 

ds 
motion  is  reversed,  -^  =  0,  and  for  this  we  have 


<.  =  icos-'(l--^); 


nn 

and  if  the  direction  of  rotation  is  reversed,  -^  =  0,  and  (8)  gives 


.  1  ,/,  GDk'\ 

^.=  -^cos-^l--^j; 


from  which  it  appears  that  if  v  <  ¥go  -^  r  the  motion  of  the 
centre  will  be  reversed,  but  otherwise  the  angular  motion  will 
be  reversed.    The  value  of  ^j  in  the  former  case  will  be  less  than 

^jz .  Both  motions  will  change  at  the  instant  of  greatest  elon- 
gation if  rv  =  ¥oo. 

If  the  values  of  t^  and  ^,  are  both  less  than  rt  -r-  D,  one  motion 
will  change  sign  before  the  instant  of  greatest  elongation  and 
the  other  after ;  otherwise  only  one  will  change  sign.     To  find 


448  PROBLEMS. 

the  total  variable  movement,  make  Dt  =  n,  and  (7)  and  (9) 
give 


^  =  (-4)5- 


If  (6)  reduces  to  zero  when  Dt  =  n^  the  body  would  rest  at 
the  moment  the  string  regains  its  original  length,  and  F=z  ^v, 
but  it  would  still  have  an  angular  velocity  of  oj  +  {rv  -=-  k^),  as 
shown  by  (8).  Similarly,  if  the  rotary  motion  is  destroyed  at 
that  instant,  the  linear  velocity  will  he  v  +  {Ic^oo  ~  r),  and  will 
continue  uniform.  It  may  be  shown  that  the  kinetic  energy  of 
the  moving  body  at  the  end  of  the  variable  motion  is  the  same 
as  at  the  beginning. 

It  has  not  been  attempted  to  solve  the  general  case  represented 
by  equation  (1).  It  is  evidently  intricate. — (Problem  and  solu- 
tion by  the  author  in  The  Analyst,  Jan.,  1882.) 

54.  Find  the  minimum  eccentricity  of  an  ellipse  capalle  of 
resting  in  equilibrium  on  a  perfectly  rough  inclined  plane,  in 
clination  j3. 

The  centre  of  the  ellipse  will  be  vertically  over  the  point  of 
support,  and  since  the  plane  is  tangent  to  the  ellipse,  a  vertical 
through  the  point  of  support  and  a  parallel  to  the  plane  through 
the  centre  will  make  conjugate  diameters.  Hence  the  acute 
angle  of  the  conjugate  diameters  is  90°  —  y5. 

At  the  point  bordering  on  motion,  the  potential  energy  is  a 
maximum,  and  the  major  axis  will  bisect  the  acute  angle  of  the 
conjugate  diameters  ;  hence  the  positive  angles  made  by  the 
conjugate  diameters  with  the  major  axis  will  be  ^  =  45°  —  ^fi, 
6'  =  135°  +  i/3. 

The  condition  for  conjugate  diameters  is 

a^  sin  6  sin  6'  -f  h*  cos  6  cos  6'  =  0, 

which,  by  substituting  the  preceding  values,  gives 

a»  sin^(45°  -  i^)  -  ¥  cos»(45°  -  J/J)  =  0  ; 


FLUIDS.  449 

which,  reduced,  gives 

|/(^')=^=|/(iTlr^)- 

(Math.   Visitor,  Jan.,  1879.) 

FLUIDS. 

55.  A  sphere  4  inches  in  diameter,  specific  gravity  0.2,  is 
placed  10  feet  under  water »  If  left  free  to  move,  what  will  be  its 
velocity  at  the  surface  of  the  water,  and  what  will  he  the  maxi- 
mum height  it  will  attain. 

Let  r  =  I  —  radius  of  the  sphere,  />  =  ^  =  its  specific  gravity, 
A  =  10  feet,  V  the  velocity  acquired  in  ascending  a  distance  x, 
Fthe  velocity  at  the  surface  of  the  water,  h  =  the  resistance. 
For  the  motion  of  the  sphere, 

vdv       /I       A      ,  , 
dx 


Putting  ^^i-lj  =  ^'. 


,  vdv 

dx  —  ~ 


vdv 


whence, 


f* ,        f  ^     vdv 

F«=|'(l-e-2^*), 
=  ^^  r      nearly. 


The  required  height  is 

I^«        1  _  n 


\-p 


%kp  "''"•^y- 


450  PROBLEMS, 

The  value  of  Tc  reduced  from  Newton's  Principia,  Book  3, 
3 
Prop.  38,  is  — — ,  whicli  gives 


F=[^/'^(l~p)]^  =  |^/6^, 


and 


A  =  TT-  =  44  inches  nearly. 
2^ 


(Math,  Visitor,  Jan.,  1879.) 

56.  The  first  of  two  cashs  contains  a  gallons  of  tvine,  and  the 
second  b  gallons  of  water  j  c  gallons  were  drawn  from  the  second 
cash,  and  then  c  gallons  were  drawn  from  the  first  cask  and 
poured  into  ilie  second,  and  the  deficiency  in  the  first  supplied  hy 
c  gallons  of  water  ;  c  gallons  were  then  drawn  from  the  first  cash, 
and  c  gallons  from  the  second,  and  poured  into  the  first,  and  tlie 
deficiency  in  the  second  cash  supplied  hy  c  gallons  of  tvine.  Re- 
quired the  quantity  of  wine  in  each  cash  after  n  such  operations 
as  that  described  above. 

Let  u^  and  v„  represent  the  wine  in  the  first  and  second  casks 
respectively  at  the  end  of  the  ^th  operation ;  the  quantities  of 
wine  in  each  cask  at  the  successive  stages  of  the  {n  +  l)th  opera- 
tion are 

Whence 


FLUIDS. 


r..=  (l-iy..+i(l-0..  +  .. 


451 

(2) 


Also 


Eliminating  v„  from  (1)  and  (2), 
Eliminating  v^^i  from  (3)  and  (5), 

"■■.-[(>-iy-iK'-9>"-('-:J(-'J- 

Let  ri,  T-j,  be  the  roots  of  the  equation 

The  solution  of  (6)  is,  (Hymer's  Finite  Differences,  pp.54,  55), 


u,=  C,r';  +  C^r'^  + 


-('-!) 


1-1 


,-)'-:i-('-i)H-a'(-i) 


Let  the  last  expression  =  S.     From  (1),  u^  =  a, 

u^  ^  {a  -  cY  ^c^  ^     c,-hC,  =  a-S,  r,C,  +  r,C,  =  u,  -  S ; 
a  0 


452 

PROBLEMS. 

whence 

r,  -  u 

• 

(y  ^  ^M  -  S)  -Ui  +  S 

(7) 


r,  —  ra 
and 

Eliminating  u^  from  (1)  and  (2),  and  tf„+i  from  this  and  (4), 

'■.-[(-.-)'4<'-f)']— ('-:-)X-0"- 
='['-(-:-)'-a.  • 

whence 

From  (2), 

t;o  =  0,     Vi  =  c(2-^^\, 
therefore 

{Math.  Visitor,  Jan.,  1880.) 

57.  ^  servant  draws  off  a  gallon  a  day  for  20  days,  from  a 
cask  holding  10  gallons  of  wine,  adding  each  time  a  gallon  of 
water  to  the  cask.  He  then  draws  off  20  gallons  more,  adding 
as  taken,  a  gallon  of  wine  to  the  cask  for  each  gallon  drawn. 
How  much  water  remains  in  the  cask  9 

Put  10  gallons  =  a,  1  gallon  =  5,  20  =  t,  and  let  u^  =  the 
number  of  gallons  of  water  in  the  cask  at  the  end  of  the  a;th 
operation.     Then  we  have 


(^ ju^  -h  b  =  Wx+i,  (1) 


FLUIDS.  453 

an  equation  in  Finite  Differences.     Integrating  (1), 

'a  —  J' 


-  =  <"-^)'-«-  ,(«) 


When 

a;  =  0,     w^  =  0  ;     .-.   C  =  —  a. 


.("-^-'>' 


)='■ 


by  the  first  condition. 

Let  V,  =  the  quantity  of  water  in  the  cask  at  the  end  of  the 
/th  operation  under  the  second  condition  ;  then, 

a  —  l 
a 
Integrating, 

When 

^  =  0,    Vo  =  c  ; 

/a  -  by  /or  -  (a  -  h)'\ 
When 

v,  =  10(.9^<'-.9*«). 

{The  Wittenberger,  Jan.,  1880.) 

68.  A  piston,  tveight  w,  is  dropped  into  the  end  of  a  vertical 
cylinder  filled  icith  air,  length  I ;  how  far  will  the  piston  descend, 
assuming  no  friction  nor  escape  of  air,  nor  heat  from  the  com- 
pressed air  ? 


454  PROBLEMS. 

There  being  no  escape  of  heat,  the  law  of  pressure  will  be  ex- 
pressed by 

pv*^  —  constant  =  v'v'^,  (1) 

where  ^'  =  the  initial  pressure  of  the  atmosphere  =  15  lbs. 
nearly  ;  v'  =  volume  of  the  cylinder  =  al,  if  a  is  the  area  of  the 
base  ;  f  =  the  pressure  within  the  cylinder  when  the  weight  has 
descended  a  distance  x  \  a{l  —  x)  =  the  volume  when  the  press- 
ure is  J9  ;  Jc  =  1.408  ;  then,  from  (1), 


and 


^-  (l-x)^' 


Integrating,  observing  that  for  x  =  0,  v  =  0, 


pa\  2ap'g    fl{l  -  a;)^'^  -  V 

w)^^'^  wfjc-l)\      (l-x)^-' 


At  the  end  of  the  downward  movement  v  =  0  ;  therefore 
x{w  +  pa){l  -  x)"-^  +  ^^  (I  -  xr-^  =  f^l, 


from  which  x  may  be  found  by  trial  after  numerical  values  have 
been  substituted  for  the  known  quantities. 

59.  If  each  of  n  vessels  closely  connected  in  circuit  contains  a 
different  liquid,  each  q  gallons,  and  the  liquids  circulate  iy  flow- 
ing ujiiformly  in  one  direction  at  the  rate  of  2i,  gallons  per  min- 
ute, mixing  uniformly,  how  much  of  each  liquid  toill  there  'be  in 
any  one  of  the  vessels  at  the  end  of  the  time  t. 

Let  the  vessels  be  numbered  in  the  natural  order  1,  2,  3  .  .  .  n; 
let  X  denote  any  particular  liquid  and  Xi,  x^,  x^,  .  .  .  x„,  the 
quantity  of  it  in  vessels  1,  ;^,  etc,  respectively  at  time  t*     cidt  =  the 


FLUIDS. 


455 


amount  flowing  out  (and   in)   each   instant,  of  which  —  adt, 

—  adt .  .  .  -^  adt.  will  be  of  the  liquid  denoted  by  x.     In  vessel 
q  q 

X  X 

1,  -  adt  flows  out,  and  -^  adt  flows  in  from  the  »th  vessel,  and 

q  ^ 

the  difference  will  be  an  element  of  the  decrease  of  the  x  liquid  ; 


dx.  =  —  adt adt. 


Letting  -  =  r,  and  dividing  by  dt,  we  have 


Xi  —X, 


or 


rdxx 
'di 

rdx. 


Similarly  for  the  others, 
rdXi 


—  ar,  =  —x^ 


rdx-i  _ 

-dt"  -  ^3  -  -  a:, 


rdx, 
dt 


-x^=  -X, 


{a) 


(« 


Differentiating  n  —  1  times,  we  have 


rd'Xi 

d'^-'x,         dr-'x^  ] 

dt^- 

dt*-^  ~       dt''-' 

rd^-'x. 

6?"-*a;,           d*-^x. 

df--' 

df^-^  ~       dt"-* 

rdx^ 
dt 

-x,=  -x^ 

(c) 


456  PROBLEMS. 

Similarly, 

df"        dF^^  "  ~  W^' 

rd*-^Xt  _  ^""'^a  _  _  ^"^ 


rdXi 
and  finally, 


dt        ^'  ~  ~^' ' 


df*         ~dF^  ~      ~dF^ ' 


dt 


X„_i  Xn-2 


Multiplying  the  successive  equations  in  (c)  by  the  successive 
terms  of  the  development  of  the  binomial  (r  —1)""^  and  add- 
ing results,  we  have 

r"d"Xi      nr"~'^d"-'^Xi     n(n  —  1)  r^'-^d^-^  nrdxi 

~W  df^-'      "^        ^  dF^ '-^  ~df  "^  ^' 


—    __    . 11    _1_    Iwi    __     I   1     1*  J: ~ L      ^ 


The  second  member  of  this  equation  is  one  degree  less  than 
the  first,  and  since  this  is  true  for  all  values  of  n,  it  will  be 


FLUIDS.  467 

true  if  we  reduce  the  exponent  and  subscript  still  further  by  1. 
This  process  ultimately  gives  for  the  second  member. 


.  rdx^ 


hence  we  have 


r^d^x^  _  wr'-W*'*^;,      n(n—  i)  r'*~*d'*~*Xi 

11  f-  dt*-'       "^        [2  W^  '  •  • 

n{n  —  1)  r^d*Xi  .  nrdx^  _  ^ 
^      (2       W     IT  ~ 

Adding    =F  1  to  both   members,  the  characteristic  equation 
becomes  (Price's  Infinitesimal  Calculus,  Vol.  II.,  p.  634), 


r'p*  -  nr-'/3*-' ±  wryS  zp  1  =  =F  1, 

or 

(r/J-l)''==Fl; 


^^1±V^ 


which  may  be  written 


^=ij^4=^-,  (^ 


where  the  exponent  n,  of  —  1,  is  simply  for  the  purpose  of 
determining  the  sign  of  this  term.  If  n  be  even  (--  1)"  will  be 
4-  1,  and  there  will  be  two  real  roots  +  1  and  —  1 ;  if  w  be  odd 
there  will  be  one  real  root  of  V—  1  =  —  !•  The  other  roots 
will  be  imaginary  and  in  pairs.  The  number  of  values  of  /3  will 
equal  the  degree  of  the  equation,  one  value  being  zero,  and 
letting  these  values  be  b^  ^„  b^  .  ,  ,  b^.i,  the  integral  becomes 
(Price's  Infinitesimal  Calculus,  Vol.  IL,  eq.  (107),  p.  637), 

X,  =  ae*.«  +  C^^f-  + +  C^e^nt^  (e) 

and  similar  expressions  for  Xt,  x^y  etc. 


458 


PROBLEMS. 


To  find  the  constants  of  integration  we  have  for  t  =  0,  x^  =  qf 

doc 

-~  =  a,  from  {a)  since,  initially,  x^  =  0, 

d^xn     _  fa  dxi      a  dx„\'~\     _  «'  _  r^  _  ^\ 
^^'  Jt=o~  \q'dt~q  'dtJjt:.Q~  q  ~q* 


^xn 


...  q=C,  +  C,+  C,    . 


i ,  etc. 


•       •       '^m 


.       .       5„a 


(/) 


:=.  JrCi  +  l\C, 


hlC. 


and  finally 


^_-  =  Jr^. +  s; 


h".C, 


from  which  the  constants  of  integration  can  be  found 
As  a  special  case  let  n  =  3,  then  from  (d). 


13  =  0,     A(3_V-3),     andJ(3  +  V-3). 


"We  then  have  from  (c), 

a?!  =  Ci  +  C^e 
and  similarly. 


^C3-v^-8)«  ^     ^(3^v^-3)« 


2« 


JjJj   ^X      I"    -^2^ 


2g 


(3_v/-3)< 


+  A. 


^(8+v^-8)« 


.2* 


(5') 


The  sum  of  these,  ox  Xi-{-x^  +  Xi  =  q  at  any  time  t 


FLUIDS.  459 

Also  by  the  process  shown  in  (/),  we  have  for  /  =  0,  x,  =  5', 
dx^  d'Xi      a'  -     dx^  d'x^  2fir'  _ 

W  =  "'  -dF=g'   ^•=*''    01=-"'  -dF=-Y'  ^^  =  ^' 
dx^  _       d^x-i  _  (f 
~dt~'W~q' 

Hence,  to  find  the  constants  in  equation  (g),  we  have, 
for  Xi , 

q  z=  Ci  +  Ci  +  C3, 


i=k^^  - V^3r6;  +  J,  (3  +  V-  3)'C. ; 

tor  x„ 

0  =  J,  +  J,  +  A„ 


-a  =  ^(3-V-3M,  +  ^(3  +V-3M,, 

and  for  .Tj, 

0  =  Z?,  +  5,  +  5,, 

0  =  ^  (3  -  V-^3)B,  +  ^  (3  +  V~3)5„ 


__^,(3-V-3r5.  +  j^(3+V-3)'5. 

These  give 

C,  =  C.  =  Ca  =  iq- 

A,  =  iq,     ^,  =  -  iq(l  +  V^,     ^3  =  -  4^(1  -  V~^)' 

B,  =  iq,     B,  =  -  iq{l  -  V^=^),     ^3  =  -  lg(l  +  V^=^). 


460 

These  in  (g)  give 


PROBLEMS. 


=  -iyf  1  +  2«  =9  cos  -^ —  j ; 


=  i?[2  - 


8a< 

2e  ^Q 


cos— r- HV3  sin 

2q 


.3  =  i,[2-2.l^(cos4^^-V3sin4%^^^^^ 

In  a  similar  manner,  if  y  is  the  liquid  in  the  second  vessel, 
the  quantity  of  it  in  vessel  2  at  the  end  of  time  t  will  be  ?/i  =  x^, 
above ;  and  similarly  for  the  others. 

60.   To  find  the  velocity  of  an  ice  boat. 

Let  AB  represent  the  track  of  the  boat,  BD  the  position  of 
the  sail,  6  =  DBA  and  ITG'  the  direction  of  the  wind,  which 

we  will  assume  to  be 
^  normal    to    the    sail. 

When  the  boat  ad- 
vances to  C,  the  posi- 
tion of  the  sail  will  be 
CE,  If  V  be  the  ve- 
locity of  the  boat,  pro- 
portional to  BC,  and 
V  the  velocity  of  the  wind  relatively  to  the  earth,  then  will 


-^c 


V  —V  Bind 


be  the  velocity  relatively  to  the  sail,  since  the  wind  passing  any 
point  as  B  must  travel  a  distance  BH  before  coming  in  contact 
with  the  sail.  The  pressure  of  the  wind  is  assumed  to  vary  as 
the  square  of  the  velocity  relatively  to  the  surface  pressed,  and 
if  M  be  the  mass  of  the  boat  and  sail,  B  a  constant  depending 


PROBLEMS.  461 

upon  the  size  of  the  sail  and  the  unit  of  velocity,  and  neglecting 
all  resistances,  we  have 


d 


/    V  ds\ 

VsirT^  ~di) 


where  A  =  — jr^ — .     Integrating,  making  F  =  0  for  ^  =  0, 


ds  _  ^  _      sin^ 

from  which  it  will  be  seen  that  V  increases  as  t  increases  in- 
definitely, the  limit  of  V  being 

F= 


sin^ 


and  this  increases  as  6  decreases,  from  which  it  appears  that, 
according  to  the  above  hypothesis,  the  boat  might  attain  an 
immense  velocity  for  a  small  velocity  of  the  wind.  The  smaller 
the  angle  of  the  sail  with  the  track  of  the  boat,  the  greater  the 
ultimate  velocity  df  the  latter,  there  being  no  resistances. 

But  the  resistances  may  be  considerable.  The  coefficient  of 
friction  on  the  ice  may  possibly  be  as  low  as  0.04.  Should  the 
air  move  away  from  behind  the  sail  with  the  velocity  of  the 
wind,  or  even  with  the  velocity  Fsin  ^,  no  resistance  would  be 
offered  by  the  air ;  but  such  will  not  be  the  case.  The  mast, 
sail,  and  other  parts  will  be  opposed  by  the  resistance  of  the  air 
as  they  move  through  it,  but  it  is  difficult  to  determine  the 
exact  amount.  General  Tower,  in  Van  Nostrand's  Engineering 
Magazine,  January,  1880,  pp.  83,  84,  according  to  an  example 
of  probable  conditions,  concluded  that  the  maximum  velocity  of 
an  ice  boat  might  exceed  twice  that  of  the  wind. 


462 


PEOBLEMS. 


61.  Determine  the  path  of  a  rotating  tody  projected  into  a 
resisting  medium. 

A  general  solution  of  this  problem  has  not  been  obtained,  but 
certain  qualitative  results  may  be  determined  without  finding 

the  quantitative.  Let  the  body  be 
a  sphere  rotating  about  a  vertical 
axis,  the  centre  moving  in  a  hori- 
zontal plane.  Let  a?  be  the  angu- 
lar velocity,  and  v  the  velocity  of 
the  centre ;  then  vrill  OK,  the 
distance  of  the  spontaneous  axis  K 
from  the  centre  be,  equation  (199), 


The  combined  motion  of  rotation  and  translation  of  the  body 
at  any  instant  being  considered  the  same  as  that  of  the  entire 
body  rotating  an  infinitesimal  amount  about  the  axis  K,  the 
quadrants  A  and  D  will  move  with  equal  velocities  exceeding 
those  of  B  and  C,  The  resistance  of  the  medium ^to  a  body 
moving  normally  agamst  it  will  vary  as  some  power  of  the 
velocity  of  the  body,  and  in  this  case  may  be  considered  as  pro- 
portional to  the  same.  The  quadrant  A  moves  against  the 
medium  with  a  greater  velocity  than  B,  hence  the  pressure  on 
that  quadrant  will  be  greater,  while  the  velocity  of  the  quadrant 
D  moving  away  from  the  medium,  exceeds  that  of  C,  There- 
fore the  pressure  of  the  medium  will  be  greatest  on  quadrant  A, 
next  B,  then  (7,  and  least  on  D.  The  resultant  of  these  press- 
ures will  not  be  zero,  and  generally  not  parallel  to  OG.  Let  R 
be  the  resultant,  the  component  of  which,  parallel  to  OG,  will 
be  the  pressure  directly  opposing  the  motion,  and  aJ,  normal  to 
OG,  the  pressure  which  will  deflect  it  from  its  initial  direction. 
Neglecting  friction,  the  path  will  be  a  curve  convex  towards 
the  quadrant  of  greatest  pressure,  and  will  be  more  nearly  a 
right  line  as  ^is  more  remote,  or  the  more  v  exceeds  go.  Still, 
neglecting  friction,  the  rotary  motion  will  be  constant,  while 
the  velocity  of  the  centre  will  be  diminished  ;  hence  the  curva- 
ture of  the  path  will  increase  with  the  distance  traveled. 

The  friction  between  the  medium  and  body  tends  directly  to 


PROBLEMS.  463 

diminish  the  rotation,  but  if  the  sum  of  the  components  of  the 
frictional  resistances  resolved  in  reference  to  two  rectangular 
planes  be  not  zero,  there  will  be  a  resultant.  The  friction  will 
be  greatest  where  the  pressure  is  greatest,  as  at  A,  and  act  tan- 
gentially  to  the  surface.  Let  F  be  the  resultant ;  it  will  be 
equal  to  a  couple,  and  an  equal  parallel  force  at  the  centre,  the 
former  of  which  reduces  the  rotation ;  and  of  the  latter,  that 
component  which  is  parallel  to  OG  directly  opposes  the  motion, 
and  that  which  is  normal  to  OG  tends  to  deflect  the  path  in  the 
direction  OB,  opposite  to  that  produced  by  pressure  only.  If  the 
body  be  comparatively  smooth  and  the  medium  rare,  the  friction 
will  be  only  a  fractional  part  of  the  pressure,  and  the  resultant 
friction  will  be  only  a  very  small  part  of  the  entire  friction,  in 
which  case  the  direction  of  the  curvature  of  the  path  will  be  de- 
termined by  the  Tesnltajit  pressure,  but  the  amount  of  curvature 
will  be  diminished  by  the  friction.  This  case  is  illustrated  by 
a  rotating  sphere  projected  into  air. 

But  if  the  body  be  rough  and  the  medium  dense,  frictional 
resistance  might  exceed  the  pressure,  in  which  case  the  direc- 
tion of  curvature  would  be  determined  by  the  resultant  friction, 
the  amount  being  modified  by  pressure.  This  could  be  illus- 
trated by  a  wheel  with  flat  vanes  rotating  about  its  axis,  placed 
vertical  and  pushed  along  in  water.  Or  still  more  strikingly,  if 
a  rough  cylinder  rotating  about  a  vertical  axis  be  pushed  into 
a  bank  of  earth,  the  tendency  to  a  lateral  motion  might  be 
almost  entirely  dependent  upon  the  friction. 


APPENDIX     II 


THE  POTENTIAL. 
80 


THE   POTENTIAL. 

The  term  *' potential  function,"  or  aimp\y  t?ie  pote7ttial,  os 
used  by  Gauss  and  subsequent  writers,  is  applied  to  a  certain 
expression  appearing  in  certain  investigations  involving  forces 
depending  upon  some  function  of  the  distance  between  the 
bodies.  It  is  of  value  in  the  higher  investigations  of  the  theory 
of  attraction,  hydromechanics,  electricity,  magnetism,  and  heat. 

Before  defining  it  definitely,  take  an  example.  Let  m  and  7n' 
be  the  masses  respectively  of  two  particles,  the  place  of  m  being 
X,  y,  Zf  of  rti,  x ,  y'  z ,  the  distance  between  them  r,  and  f{r) 
the  law  of  the  mutual  attraction  or  repulsion.  Then  will  the 
stress  between  them  be 

P=^mmf{r\  (a) 

—  being  attractive  and  +  repulsive.     The  axial  components 
will  be 

Pcosa  =  Jr=ww7(r)^^^^,     Y=mm'f(r)y-^, 

and  for  the  distance  between  them, 

r«  =  (x'  -  xy  4-  (;/'  -  yy  +  («'  -  zy  ;  (c) 

from  which  we  have  for  the  partial  differential  coefficients,  x,  y, 
z,  being  considered  fixed, 

rdr  —  (x'  —  x)dx\     rdr  =  (y'  —  y)dy',     rdr  =  {z  —  z)dz.     (d) 

Let/(r)  be  considered  as  the  differential  coefficient  of  some 
other  function  of  r,  so  that 


467 


468  •  APPENDIX  II. 

then  will  equations  (d)  and  (e)  reduce  {b)  to 

^        ,    dF{r)       ,^         ,     dF{r)       „         ,     dF{r) 
X  =  mm  — ^-V  9     V=mm  — 7-V  i     Z  =  mm  — H-^ . 
dx  dy  dz 

If  ma,  ma,  etc.,  be  the  masses  of  other  particles  distant  r,,  rj, 
etc.,  from  m',  then  we  have,  m  being  typical  of  mj,  m^,  etc., 

^     ^^^Vtz^^  d:EmF(r)  d^mFyr) 

Let 

^mi^(r)  =  F,      '  (^) 

then  we  have 

'=-'©.  '-"•■Cv)-  '"•'(-£)■■   « 

the  parentheses  indicating  partial  differential  coefficients.  The 
function  Fis  called  the  potential,  and  may  be  defined  by  inter- 
preting equation  {g).  When  found,  the  axial  components  of  the 
stress  appear  as  partial  differential  coefficients  of  V  regarded  as 
a  function  of  the  coordinates  of  the  particle.  It  is  rarely  used 
in  this  general  form,  but  is  confined  to  the  cases  where  the  law 
of  the  force  is  that  of  the  inverse  squares — the  law  most  common 
in  nature.  Let  the  system  of  particles  7)1  be  continuous,  form- 
ing a  solid,  and  the  force  attractive,  then  we  have 

and  if  <y  be  the  density  at  the  place  x,  y,  z,  then 
dm'=  6dxdydz\ 


that  is,  The  potektial  is  sum  of  the  quotients  of  all  the  element- 
ary masses  divided  by  their  distance  from  the  attracted  particle. 


THE  POTENTIAL.  469 

Again,  if  the  mass  of  the  attracted  particle  be  unity,  nnd  that 
of  the  attracting  particle  be  m,  then  at  any  distance  r  we  have 
the  stress, 

and  hence  an  element  of  the  work  done  upon  the  unit  mass  at 
that  f)oint  in  being  moved  over  the  space  dr  will  be 


dw  = idr  \ 


to 


=  -\:>=r'  0) 


and  similarly  for  any  number  of  particles  forming  a  continuous 
body  ; 


which  is  the  same  form  as  that  above,  hence,  also. 

The  potential  is  the  energy  acquired  hy  a  unit  mass  in  fall- 
ing from  infinity  under  the  attraction  of  a  given  body  to  a  dis- 
tance r. 

Again,  m  may  represent  any  quantity  of  action,  either  attrac- 
tive or  repulsive,  as  in  magnetism  or  electricity,  in  which  the 
law  of  action  is  that  of  the  inverse  squares  and  product  of  their 
quantities. 
From  equation  (c)  we  have  the  partial  derivative 

,  7  ,v         rdr 


and  from  (i),  considering  F  as  a  function  of  r, 

(dy\  _  _  f  f  f  ^{p^  —x)dxdydz 
'''  \dx')  - 


470  APPENDIX  II. 

similarly, 

(£f)=lll['-'^?^4>*- 

and  adding 

O-Q- (£■?)=»  ^       <" 

which  theorem  was  discovered  by  La  Place.  It  is  not  general, 
however,  for  it  is  found  to  fail  when  the  particle  is  one  of  the 
particles  of  the  attracting  mass ;  but  it  is  correct  when  the 
particle  attracted  is  external  to  the  attracting  body. 

Examples. 

1 .  To  find  the  ^potential  of  a  slender  uniform  rod,  length  a, 
density  d,  and  section  s,  upon  cm  external  particle  m! . 

Take  the  origin  at  one  end  of  the  rod,  x,  along  the  rod,  and 
X ,  y\  the  position  of  the  particle  m! .     Then 

V  —      — =  6s  log ,  ^^  ,  ,, ^rf^  .  (m) 

Jo  V(x'  -  xy  +  y"  -  «  +  (^    +  ^) 

Hence  the  attractive  forces  parallel  and  normal  to  the  rod 
will  be  respectively, 

X  =  —  m  (  ^-^     =  mds       ,  ,  . 

\dx  J  V-Vy'^  +  {a  -  xj      ^y""  +  x'^A 

2.  To  find  the  potential  of  a  thin,  homogeneous,  spherical  shell 
upon  an  external  particle. 


THE  POTENTIAL.  471 

Let  a  —  radius,  S  =  density,  da  —  thickness,  p  —  distance  of 
particle  from  the  centre  of  the  shell.  Using  polar  coordinates, 
origin  at  the  centre  of  the  shell,  (0,  0,  p)  the  place  of  m , 
6  =  polar  distance  of  element  of  m  from  where  p  pierces  the 
shell,  (p  =  longitude,  initial  at  any  poiat,  then 

dm  =  6 da  •  a  sin  6dcp  -  add, 

r^  —  a^  —  Hap  cos  0  -r  p' ; 

.17      ^7        J'  f^'  sin  OdOdcp 

,',    y  =  6da  •  a' \  /-5      o  -n~, — •si.* 

Jo  Jo   (a' —  2ap  cos  ^  +  p-)* 

27rSada  /  ,      ^  /a   .     2^*1'' 

= i  a*  —  2ap  cos  ^  +  p  )        , 

^TtSada  /     ,  /  s\  ,  N 

[a  -h  p-(a-  p)J,  («) 


or. 


P 
^ndada 


a  -\-  p-{p-a)\  (o) 


For  an  external  particle  p>  a,  hence  the  last  equation  is  the 
correct  form  for  this  case,  and  the  former  gives  the  potential 
for  a  particle  internal  to  the  shell.     For  the  latter, 

and  for  a  concentric  shell  of  finite  thickness, 

•'•   \dp)  -      p«  ' 

which  is  the  force  along  the  line  p  ;  hence, 

Tlie  attraction  of  a  spherical,   homogeneous  shell  upon  an 


472  APPENDIX   II. 

exterior  particle  is  the  same  as  if  the  mass  of  the  shell  were  con- 
densed into  a  particle  at  the  centre. 
Equation  {n)  becomes 


Y  =  4:7tdada ; 


hence,  The  attraction  of  a  spherical  homogeneous  shell  upon  a 
particle  within  it  is  zero. 


X 


47'^+^x^i=  jc;ilt*''-f^-  ^c-^-'^*jfx-'-y_'^  ^u-fk. 


trC^ 


'V-f  ^^^     ■+    t^x-j-  i) 


tlrtii«-<_  .• 


/->« 


^     'ys^*  -  A  *-  ^^i  »w  -// 


■h^ 


r-u^ 


iZJ*^ 


-  :i: 


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V 

AUG  »&   1946 

^H~ 

AUG  29  1046 

^V 

^m. 

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jHJl 

^^ 

1 

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IH 

1^1 

^H 

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LD  21-100m-7,'40  (6936s) 


IV!30<5889 

W7 


THE  UNIVERSITY  OF  CAUFORNIA  UBRARY 


